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# NCERT Exemplar for Class 11 Maths Chapter 3 - Trigonometric Functions (Book Solutions)

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## NCERT Exemplar for Class 11 Maths - Trigonometric Functions - Free PDF Download

Free PDF download of NCERT Exemplar for Class 11 Maths Chapter 3 - Trigonometric Functions solved by expert Maths teachers of Vedantu as per NCERT (CBSE) Book guidelines. All Chapter 3 - Trigonometric Functions exercise questions with solutions to help you to revise the complete syllabus and score more marks in your examinations.

Trigonometry comes with different real-time applications. This subject is popular for resolving distance and height problems. This specific chapter offers a complete introduction to the basic properties as well as evaluates trigonometric questions and functions. Vedantu provides answers to the students according to  NCERT Exemplar for Class 11 Maths Chapter 3 - Trigonometric.

Competitive Exams after 12th Science

## Access NCERT Exemplar Solutions for CBSE Class 11 Mathematics Chapter 3: Trigonometric Functions (Examples, Easy Methods and Step by Step Solutions)

Examples

Short Answer Type

Example 1: A circular wire of radius $3cm$ is cut and bent so as to lie along the circumference of a hoop whose radius is  $48cm$ . Find the angle in degrees which is subtended at the centre of hoop.

Ans: Given that, radius of circular wire = $3cm$

When it is cut then its length becomes $2\pi \times 3$ = $6\pi$

Again, it is being placed along a circular hoop of radius $48cm$

The length $\left( s \right)$ of the arc = $6\pi$

Radius of circle, $r = 48cm$

Therefore, the angle $\theta$ (in radian) subtended by the arc at the centre of circle is given by

$\Rightarrow \theta = \dfrac{{Arc}}{{Radius}}$

$\Rightarrow \theta = \dfrac{{6\pi }}{{48}}$

$\Rightarrow \theta = \dfrac{\pi }{8}$

$\Rightarrow \theta = 22.5^\circ$

Example 2: If $A = {\cos ^2}\theta + {\sin ^4}\theta$ for all values of $\theta$, then prove that $\dfrac{3}{4} \leqslant A \leqslant 1$

Ans: Given that, $A = {\cos ^2}\theta + {\sin ^4}\theta$

$\Rightarrow A = {\cos ^2}\theta + {\sin ^2}\theta {\sin ^2}\theta \leqslant {\cos ^2}\theta + {\sin ^2}\theta$

We know that ${\sin ^2}\theta + {\cos ^2}\theta = 1$. Hence, we get

$\Rightarrow A = {\cos ^2}\theta + {\sin ^2}\theta {\sin ^2}\theta \leqslant 1.......\left( i \right)$

Now, $A = {\cos ^2}\theta + {\sin ^4}\theta$

We know that ${\sin ^2}\theta + {\cos ^2}\theta = 1$. So, we can write above written equation as,

$\Rightarrow A = \left( {1 - {{\sin }^2}\theta } \right) + {\sin ^4}\theta$

$\Rightarrow A = {\sin ^4}\theta - {\sin ^2}\theta + 1$

Add $\dfrac{1}{4}$ from RHS and subtract $\dfrac{1}{4}$ from the RHS

$\Rightarrow A = {\sin ^4}\theta - {\sin ^2}\theta + \dfrac{1}{4} + 1 - \dfrac{1}{4}$

$\Rightarrow A = {\left( {{{\sin }^2}\theta - \dfrac{1}{2}} \right)^2} + \dfrac{{4 - 1}}{4}$

$\Rightarrow A = {\left( {{{\sin }^2}\theta - \dfrac{1}{2}} \right)^2} + \dfrac{3}{4}$

Therefore, $A = {\left( {{{\sin }^2}\theta - \dfrac{1}{2}} \right)^2} + \dfrac{3}{4} \geqslant \dfrac{3}{4}......\left( {ii} \right)$

Thus, from equation $\left( i \right)$ and $\left( {ii} \right)$, we get

$\Rightarrow \dfrac{3}{4} \leqslant A \leqslant 1$

Example 3: Find the value of $\sqrt 3 \cos ec20^\circ - \sec 20^\circ$

Ans: We have, $\sqrt 3 \cos ec20^\circ - \sec 20^\circ$

We can also write it as,

$\Rightarrow \dfrac{{\sqrt 3 }}{{\sin 20^\circ }} - \dfrac{1}{{\cos 20^\circ }}$

$\Rightarrow \dfrac{{\sqrt 3 \cos 20^\circ - \sin 20^\circ }}{{\sin 20^\circ \cos 20^\circ }}$

Multiply and divide numerator by $2$

$\Rightarrow \dfrac{{2\left( {\dfrac{{\sqrt 3 }}{2}\cos 20^\circ - \dfrac{1}{2}\sin 20^\circ } \right)}}{{\sin 20^\circ \cos 20^\circ }}$

We know that $\dfrac{{\sqrt 3 }}{2} = \sin 60^\circ$ and $\dfrac{1}{2} = \cos 60^\circ$. So, we can write above-written expression as,

$\Rightarrow \dfrac{{2\left( {\sin 60^\circ \cos 20^\circ - \cos 60^\circ \sin 20^\circ } \right)}}{{\sin 20^\circ \cos 20^\circ }}$

We know that $\sin \left( {A - B} \right) = \sin A\cos B - \cos A\sin B$. Therefore, we get

$\Rightarrow \dfrac{{2\sin \left( {60^\circ - 20^\circ } \right)}}{{\sin 20^\circ \cos 20^\circ }}$

Multiply and divide the above-written expression by $2$

$\Rightarrow \dfrac{{2 \times 2\sin \left( {60^\circ - 20^\circ } \right)}}{{2\sin 20^\circ \cos 20^\circ }}$

We know that $2\sin x\cos x = \sin 2x$. Therefore, we get

$\Rightarrow \dfrac{{4\sin \left( {60^\circ - 20^\circ } \right)}}{{\sin 2 \times 20^\circ }}$

$\Rightarrow \dfrac{{4\sin 40^\circ }}{{\sin 40^\circ }}$

$\Rightarrow 4$

Example 4: If $\theta$ lies in the second quadrant, then show that $\sqrt {\dfrac{{1 - \sin \theta }}{{1 + \sin \theta }}} + \sqrt {\dfrac{{1 + \sin \theta }}{{1 - \sin \theta }}} = - 2\sec \theta$

Ans: We have, $\sqrt {\dfrac{{1 - \sin \theta }}{{1 + \sin \theta }}} + \sqrt {\dfrac{{1 + \sin \theta }}{{1 - \sin \theta }}} = - 2\sec \theta$

Take LHS,

$\Rightarrow \sqrt {\dfrac{{1 - \sin \theta }}{{1 + \sin \theta }}} + \sqrt {\dfrac{{1 + \sin \theta }}{{1 - \sin \theta }}}$

$\Rightarrow \sqrt {\dfrac{{\left( {1 - \sin \theta } \right)\left( {1 - \sin \theta } \right)}}{{\left( {1 + \sin \theta } \right)\left( {1 - \sin \theta } \right)}}} + \sqrt {\dfrac{{\left( {1 + \sin \theta } \right)\left( {1 + \sin \theta } \right)}}{{\left( {1 + \sin \theta } \right)\left( {1 + \sin \theta } \right)}}}$

$\Rightarrow \sqrt {\dfrac{{{{\left( {1 - \sin \theta } \right)}^2}}}{{\left( {{1^2} - {{\sin }^2}\theta } \right)}}} + \sqrt {\dfrac{{{{\left( {1 + \sin \theta } \right)}^2}}}{{\left( {{1^2} - {{\sin }^2}\theta } \right)}}}$

$\Rightarrow \dfrac{{\left( {1 - \sin \theta } \right)}}{{\sqrt {\left( {{1^2} - {{\sin }^2}\theta } \right)} }} + \dfrac{{\left( {1 + \sin \theta } \right)}}{{\sqrt {\left( {{1^2} - {{\sin }^2}\theta } \right)} }}$

$\Rightarrow \dfrac{{1 - \sin \theta + 1 + \sin \theta }}{{\sqrt {\left( {{1^2} - {{\sin }^2}\theta } \right)} }}$

$\Rightarrow \dfrac{2}{{\sqrt {\left( {{1^2} - {{\sin }^2}\theta } \right)} }}$

We know that ${\sin ^2}\theta + {\cos ^2}\theta = 1$. Therefore, we get

$\Rightarrow \dfrac{2}{{\sqrt {\cos \theta } }}$

$\Rightarrow \dfrac{2}{{\left| {\cos \theta } \right|}}$

(Since $\sqrt {{\alpha ^2}} = \left| \alpha \right|$ for every real number $\alpha$ )

Given that, $\theta$ lies in second quadrant and we know that in second quadrant $\cos$ is negative. Therefore, we get

$\Rightarrow \dfrac{2}{{ - \cos \theta }}$

$\Rightarrow - 2\sec \theta$

Hence proved

Example 5: Find the value of $\tan 9^\circ - \tan 27^\circ - \tan 63^\circ + \tan 81^\circ$

Ans: We have, $\tan 9^\circ - \tan 27^\circ - \tan 63^\circ + \tan 81^\circ$

$\Rightarrow \tan 9^\circ + \tan 81^\circ - \tan 27^\circ - \tan 63^\circ$

$\Rightarrow \tan 9^\circ + \tan \left( {90^\circ - 9^\circ } \right) - \tan 27^\circ - \tan \left( {90^\circ - 27^\circ } \right)$

We know that $\tan \left( {90^\circ - \theta } \right) = \cos \theta$. Therefore, we get

$\Rightarrow \tan 9^\circ + \cot 9^\circ - \tan 27^\circ - \cot 27^\circ$

$\Rightarrow \tan 9^\circ + \cot 9^\circ - \left( {\tan 27^\circ + \cot 27^\circ } \right)$

Now, we will write the above-written expression in terms of $\sin e$ and $\cos ine$.

$\Rightarrow \dfrac{{\sin 9^\circ }}{{\cos 9^\circ }} + \dfrac{{\cos 9^\circ }}{{\sin 9^\circ }} - \left( {\dfrac{{\sin 27^\circ }}{{\cos 27^\circ }} + \dfrac{{\cos 27^\circ }}{{\sin 27^\circ }}} \right)$

Take LCM

$\Rightarrow \dfrac{{\sin 9^\circ \times \sin 9^\circ + \cos 9^\circ \times \cos 9^\circ }}{{\sin 9^\circ \cos 9^\circ }} - \left( {\dfrac{{\sin 27^\circ \times \sin 27^\circ + \cos 27^\circ \times \cos 27^\circ }}{{\sin 27^\circ \cos 27^\circ }}} \right)$

$\Rightarrow \dfrac{{{{\sin }^2}9^\circ + {{\cos }^2}9^\circ }}{{\sin 9^\circ \cos 9^\circ }} - \left( {\dfrac{{{{\sin }^2}27^\circ + {{\cos }^2}27^\circ }}{{\sin 27^\circ \cos 27^\circ }}} \right)$

We know that ${\sin ^2}\theta + {\cos ^2}\theta = 1$. Therefore, we get

$\Rightarrow \dfrac{1}{{\sin 9^\circ \cos 9^\circ }} - \dfrac{1}{{\sin 27^\circ \cos 27^\circ }}$

Multiply and divide the expression by $2$

$\Rightarrow \dfrac{2}{{2\sin 9^\circ \cos 9^\circ }} - \dfrac{2}{{2\sin 27^\circ \cos 27^\circ }}$

We know that $2\sin x\cos x = \sin 2x$. Therefore, we get

$\Rightarrow \dfrac{2}{{\sin 18^\circ }} - \dfrac{2}{{\sin 54^\circ }}$

We know that $\sin 18^\circ = \dfrac{{\sqrt 5 - 1}}{4}$ and $\sin 54^\circ = \dfrac{{\sqrt 5 + 1}}{4}$. Therefore, we get

$\Rightarrow \dfrac{{2 \times 4}}{{\sqrt 5 - 1}} - \dfrac{{2 \times 4}}{{\sqrt 5 + 1}}$

$\Rightarrow \dfrac{{8\left( {\sqrt 5 + 1} \right) - 8\left( {\sqrt 5 - 1} \right)}}{{5 - 1}}$

$\Rightarrow \dfrac{{8\sqrt 5 + 8 - 8\sqrt 5 + 8}}{4}$

$\Rightarrow \dfrac{{16}}{4}$

$\Rightarrow 4$

Example 6: Prove that $\dfrac{{\sec 8\theta - 1}}{{\sec 4\theta - 1}} = \dfrac{{\tan 8\theta }}{{\tan 2\theta }}$

Ans: We have, LHS $\dfrac{{\sec 8\theta - 1}}{{\sec 4\theta - 1}}$

Write the above-written expression in terms of $\cos ine$

$\Rightarrow \dfrac{{\dfrac{1}{{\cos 8\theta }} - 1}}{{\dfrac{1}{{\cos 4\theta }} - 1}}$

$\Rightarrow \dfrac{{\dfrac{{1 - \cos 8\theta }}{{\cos 8\theta }}}}{{\dfrac{{1 - \cos 4\theta }}{{\cos 4\theta }}}}$

$\Rightarrow \dfrac{{1 - \cos 8\theta }}{{\cos 8\theta }} \times \dfrac{{\cos 4\theta }}{{1 - \cos 4\theta }}$

$\Rightarrow \dfrac{{\left( {1 - \cos 8\theta } \right)\cos 4\theta }}{{\cos 8\theta \left( {1 - \cos 4\theta } \right)}}$

We know that $1 - \cos 2x = 2{\sin ^2}x$. Therefore, we get

$\Rightarrow \dfrac{{2{{\sin }^2}4\theta \cos 4\theta }}{{\cos 8\theta 2{{\sin }^2}2\theta }}$

$\Rightarrow \dfrac{{\sin 4\theta \left( {2\sin 4\theta \cos 4\theta } \right)}}{{2\cos 8\theta {{\sin }^2}2\theta }}$

We know that $2\sin x\cos x = \sin 2x$. Therefore, we get

$\Rightarrow \dfrac{{\sin 4\theta \sin 8\theta }}{{2\cos 8\theta {{\sin }^2}2\theta }}$

We know that $2\sin x\cos x = \sin 2x$. Therefore, we get

$\Rightarrow \dfrac{{2\sin 2\theta \cos 2\theta \sin 8\theta }}{{2\cos 8\theta {{\sin }^2}2\theta }}$

On canceling common terms, we get

$\Rightarrow \dfrac{{\sin 8\theta \cos 2\theta }}{{\cos 8\theta \sin 2\theta }}$

$\Rightarrow \dfrac{{\tan 8\theta }}{{\tan 2\theta }}$

Hence proved

Example 7: Solve the equation $\sin \theta + \sin 3\theta + \sin 5\theta = 0$

Ans: We have, $\sin \theta + \sin 3\theta + \sin 5\theta = 0$

$\Rightarrow \left( {\sin 5\theta + \sin \theta } \right) + \sin 3\theta = 0$

We know that $\sin A + \sin B = 2\sin \dfrac{{A + B}}{2}\cos \dfrac{{A - B}}{2}$. Therefore, we get

$\Rightarrow \left( {2\sin \dfrac{{5\theta + \theta }}{2}\cos \dfrac{{5\theta - \theta }}{2}} \right) + \sin 3\theta = 0$

$\Rightarrow 2\sin 3\theta \cos 2\theta + \sin 3\theta = 0$

$\Rightarrow \sin 3\theta \left( {2\cos 2\theta + 1} \right) = 0$

$\Rightarrow \sin 3\theta = 0$ or $2\cos 2\theta + 1 = 0$

$\Rightarrow \sin 3\theta = 0$ or $\cos 2\theta = \dfrac{{ - 1}}{2}$

Now, $\sin 3\theta = 0$

$\Rightarrow 3\theta = n\pi ,n \in Z$

$\Rightarrow \theta = \dfrac{{n\pi }}{3},n \in Z$

And, $\cos 2\theta = \dfrac{{ - 1}}{2}$

$\Rightarrow \cos 2\theta = \cos \left( {\pi - \dfrac{\pi }{3}} \right)$

$\Rightarrow \cos 2\theta = \cos \left( {\dfrac{{3\pi - \pi }}{3}} \right)$

$\Rightarrow \cos 2\theta = \cos \left( {\dfrac{{2\pi }}{3}} \right)$

We know that if $\cos \theta = \cos \alpha$, then $\theta = 2m\pi \pm \alpha$. Therefore, we get

$\Rightarrow 2\theta = 2m\pi \pm \dfrac{{2\pi }}{3},m \in Z$

$\Rightarrow \theta = m\pi \pm \dfrac{\pi }{3},m \in Z$

Hence, the general solution of the given equation is $\theta = \dfrac{{n\pi }}{3},n \in Z$ or, $\theta = m\pi \pm \dfrac{\pi }{3},m \in Z$.

Example 8: Solve $2{\tan ^2}x + {\sec ^2}x = 2$ for $0 \leqslant x \leqslant 2\pi$

Ans: We have, $2{\tan ^2}x + {\sec ^2}x = 2$

We know that ${\sec ^2}\theta = 1 + {\tan ^2}\theta$. Therefore, we get

$\Rightarrow 2{\tan ^2}x + 1 + {\tan ^2}x = 2$

$\Rightarrow 3{\tan ^2}x = 2 - 1$

$\Rightarrow 3{\tan ^2}x = 1$

$\Rightarrow {\tan ^2}x = \dfrac{1}{3}$

$\Rightarrow {\left( {\tan x} \right)^2} = {\left( {\dfrac{1}{{\sqrt 3 }}} \right)^2}$

$\Rightarrow {\left( {\tan x} \right)^2} = {\left( {\tan \dfrac{\pi }{6}} \right)^2}$

$\Rightarrow {\tan ^2}x = {\tan ^2}\dfrac{\pi }{6}$

We know that if ${\tan ^2}\theta = {\tan ^2}\alpha$, then $\theta = n\pi \pm \alpha$, $n \in Z$.

$\Rightarrow x = n\pi \pm \dfrac{\pi }{6}$

Therefore, possible solutions are $\dfrac{\pi }{6}$, $\dfrac{{5\pi }}{6}$, $\dfrac{{7\pi }}{6}$, $\dfrac{{11\pi }}{6}$ where $0 \leqslant x \leqslant 2\pi$.

Long Answer Type

Example 9: Find the value of $\left( {1 + \cos \dfrac{\pi }{8}} \right)\left( {1 + \cos \dfrac{{3\pi }}{8}} \right)\left( {1 + \cos \dfrac{{5\pi }}{8}} \right)\left( {1 + \cos \dfrac{{7\pi }}{8}} \right)$

Ans: We have, $\left( {1 + \cos \dfrac{\pi }{8}} \right)\left( {1 + \cos \dfrac{{3\pi }}{8}} \right)\left( {1 + \cos \dfrac{{5\pi }}{8}} \right)\left( {1 + \cos \dfrac{{7\pi }}{8}} \right)$

$\Rightarrow \left( {1 + \cos \dfrac{\pi }{8}} \right)\left( {1 + \cos \dfrac{{3\pi }}{8}} \right)\left( {1 + \cos \left( {\pi - \dfrac{{3\pi }}{8}} \right)} \right)\left( {1 + \cos \left( {\pi - \dfrac{\pi }{8}} \right)} \right)$

We know that $\cos \left( {\pi - \theta } \right) = - \cos \theta$. Therefore, we get

$\Rightarrow \left( {1 + \cos \dfrac{\pi }{8}} \right)\left( {1 + \cos \dfrac{{3\pi }}{8}} \right)\left( {1 - \cos \dfrac{{3\pi }}{8}} \right)\left( {1 - \cos \dfrac{\pi }{8}} \right)$

$\Rightarrow \left( {1 - {{\cos }^2}\dfrac{\pi }{8}} \right)\left( {1 - {{\cos }^2}\dfrac{{3\pi }}{8}} \right)$

We know that ${\sin ^2}\theta = 1 - {\cos ^2}\theta$. Therefore, we get

$\Rightarrow {\sin ^2}\dfrac{\pi }{8}{\sin ^2}\dfrac{{3\pi }}{8}$

Multiply and divide the above written expression by $4$.

$\Rightarrow \dfrac{1}{4}\left( {2{{\sin }^2}\dfrac{\pi }{8}} \right)\left( {2{{\sin }^2}\dfrac{{3\pi }}{8}} \right)$

We know that $1 - \cos 2A = 2{\sin ^2}A$. Therefore, we get

$\Rightarrow \dfrac{1}{4}\left( {1 - \cos \dfrac{\pi }{4}} \right)\left( {1 - \cos \dfrac{{3\pi }}{4}} \right)$

$\Rightarrow \dfrac{1}{4}\left( {1 - \dfrac{1}{{\sqrt 2 }}} \right)\left( {1 + \dfrac{1}{{\sqrt 2 }}} \right)$

$\Rightarrow \dfrac{1}{4}\left( {{1^2} - {{\left( {\dfrac{1}{{\sqrt 2 }}} \right)}^2}} \right)$

$\Rightarrow \dfrac{1}{4}\left( {1 - \dfrac{1}{2}} \right)$

$\Rightarrow \dfrac{1}{4}\left( {\dfrac{1}{2}} \right)$

$\Rightarrow \dfrac{1}{8}$

Example 10: If $x\cos \theta = y\cos \left( {\theta + \dfrac{{2\pi }}{3}} \right) = z\cos \left( {\theta + \dfrac{{4\pi }}{3}} \right)$, then find the value of $xy + yz + zx$

Ans: Note that $xy + yz + zx = xyz\left( {\dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z}} \right)$

Given that, $x\cos \theta = y\cos \left( {\theta + \dfrac{{2\pi }}{3}} \right) = z\cos \left( {\theta + \dfrac{{4\pi }}{3}} \right)$

Let $x\cos \theta = y\cos \left( {\theta + \dfrac{{2\pi }}{3}} \right) = z\cos \left( {\theta + \dfrac{{4\pi }}{3}} \right) = k$

Then, $x = \dfrac{k}{{\cos \theta }}$, $y = \dfrac{k}{{\cos \left( {\theta + \dfrac{{2\pi }}{3}} \right)}}$ and $z = \dfrac{k}{{\cos \left( {\theta + \dfrac{{4\pi }}{3}} \right)}}$

Now, we have

$\Rightarrow \dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z} = \dfrac{1}{{\dfrac{k}{{\cos \theta }}}} + \dfrac{1}{{\dfrac{k}{{\cos \left( {\theta + \dfrac{{2\pi }}{3}} \right)}}}} + \dfrac{1}{{\dfrac{k}{{\cos \left( {\theta + \dfrac{{4\pi }}{3}} \right)}}}}$

$\Rightarrow \dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z} = \dfrac{{\cos \theta }}{k} + \dfrac{{\cos \left( {\theta + \dfrac{{2\pi }}{3}} \right)}}{k} + \dfrac{{\cos \left( {\theta + \dfrac{{4\pi }}{3}} \right)}}{k}$

$\Rightarrow \dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z} = \dfrac{1}{k}\left[ {\cos \theta + \cos \left( {\theta + \dfrac{{2\pi }}{3}} \right) + \cos \left( {\theta + \dfrac{{4\pi }}{3}} \right)} \right]$

We know that $\cos \left( {A + B} \right) = \cos A\cos B - \sin A\sin B$. Therefore, we get

$\Rightarrow \dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z} = \dfrac{1}{k}\left[ {\cos \theta + \cos \theta \cos \dfrac{{2\pi }}{3} - \sin \theta \sin \dfrac{{2\pi }}{3} + \cos \theta \cos \dfrac{{4\pi }}{3} - \sin \theta \sin \dfrac{{4\pi }}{3}} \right]$

$\Rightarrow \dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z} = \dfrac{1}{k}\left[ {\cos \theta + \cos \theta \left( {\dfrac{{ - 1}}{2}} \right) - \sin \theta \left( {\dfrac{{\sqrt 3 }}{2}} \right) + \cos \theta \left( {\dfrac{{ - 1}}{2}} \right) - \sin \theta \left( {\dfrac{{ - \sqrt 3 }}{2}} \right)} \right]$

$\Rightarrow \dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z} = \dfrac{1}{k}\left[ {\cos \theta - \dfrac{1}{2}\cos \theta - \dfrac{{\sqrt 3 }}{2}\sin \theta - \dfrac{1}{2}\cos \theta + \dfrac{{\sqrt 3 }}{2}\sin \theta } \right]$

$\Rightarrow \dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z} = \dfrac{1}{k}\left[ {\cos \theta - \cos \theta - \dfrac{{\sqrt 3 }}{2}\sin \theta + \dfrac{{\sqrt 3 }}{2}\sin \theta } \right]$

$\Rightarrow \dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z} = \dfrac{1}{k}\left( 0 \right)$

$\Rightarrow \dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z} = 0$

$\therefore xy + yz + zx = xyz\left( 0 \right)$

$\Rightarrow xy + yz + zx = 0$

Example 11: If $\alpha$ and $\beta$ are the solutions of the equation $a\tan \theta + b\sec \theta = c$, then show that $\tan \left( {\alpha + \beta } \right) = \dfrac{{2ac}}{{{a^2} - {c^2}}}$.

Ans: Given that, $a\tan \theta + b\sec \theta = c$

Now, we will write the above written equation in terms of $\sin e$ and $\cos ine$.

$\Rightarrow a\dfrac{{\sin \theta }}{{\cos \theta }} + b\dfrac{1}{{\cos \theta }} = c$

$\Rightarrow \dfrac{{a\sin \theta + b}}{{\cos \theta }} = c$

$\Rightarrow a\sin \theta + b = c\cos \theta$

We know that $\sin \theta = \dfrac{{2\tan \dfrac{\theta }{2}}}{{1 + {{\tan }^2}\dfrac{\theta }{2}}}$ and $\cos \theta = \dfrac{{1 - {{\tan }^2}\dfrac{\theta }{2}}}{{1 + {{\tan }^2}\dfrac{\theta }{2}}}$. Therefore, we get

$\Rightarrow \dfrac{{a\left( {2\tan \dfrac{\theta }{2}} \right)}}{{1 + {{\tan }^2}\dfrac{\theta }{2}}} + b = \dfrac{{c\left( {1 - {{\tan }^2}\dfrac{\theta }{2}} \right)}}{{1 + {{\tan }^2}\dfrac{\theta }{2}}}$

$\Rightarrow \dfrac{{a\left( {2\tan \dfrac{\theta }{2}} \right) + b\left( {1 + {{\tan }^2}\dfrac{\theta }{2}} \right)}}{{1 + {{\tan }^2}\dfrac{\theta }{2}}} = \dfrac{{c\left( {1 - {{\tan }^2}\dfrac{\theta }{2}} \right)}}{{1 + {{\tan }^2}\dfrac{\theta }{2}}}$

$\Rightarrow 2a\tan \dfrac{\theta }{2} + b\left( {1 + {{\tan }^2}\dfrac{\theta }{2}} \right) = c\left( {1 - {{\tan }^2}\dfrac{\theta }{2}} \right)$

$\Rightarrow 2a\tan \dfrac{\theta }{2} + b + b{\tan ^2}\dfrac{\theta }{2} = c - c{\tan ^2}\dfrac{\theta }{2}$

$\Rightarrow 2a\tan \dfrac{\theta }{2} + b{\tan ^2}\dfrac{\theta }{2} + c{\tan ^2}\dfrac{\theta }{2} + b - c = 0$

$\Rightarrow \left( {b + c} \right){\tan ^2}\dfrac{\theta }{2} + 2a\tan \dfrac{\theta }{2} + b - c = 0$

Above written equation is quadratic in $\tan \dfrac{\theta }{2}$ and hence $\tan \dfrac{\alpha }{2}$ and $\tan \dfrac{\beta }{2}$ are the roots of this equation.

We know that if the roots of the quadratic equation $a{x^2} + bx + c = 0$ are $\alpha$ and $\beta$. Then we have, $\alpha + \beta = - \dfrac{b}{a}$ and $\alpha \beta = \dfrac{c}{a}$. Therefore, we get

$\Rightarrow \tan \dfrac{\alpha }{2} + \tan \dfrac{\beta }{2} = \dfrac{{ - 2a}}{{b + c}}$ and $\tan \dfrac{\alpha }{2}\tan \dfrac{\beta }{2} = \dfrac{{b - c}}{{b + c}}$

We know that $\tan \left( {x + y} \right) = \dfrac{{\tan x + \tan y}}{{1 - \tan x\tan y}}$. Therefore, we get

$\Rightarrow \tan \left( {\dfrac{\alpha }{2} + \dfrac{\beta }{2}} \right) = \dfrac{{\tan \dfrac{\alpha }{2} + \tan \dfrac{\beta }{2}}}{{1 - \tan \dfrac{\alpha }{2}\tan \dfrac{\beta }{2}}}$

On substituting the values, we get

$\Rightarrow \tan \left( {\dfrac{\alpha }{2} + \dfrac{\beta }{2}} \right) = \dfrac{{\dfrac{{ - 2a}}{{b + c}}}}{{1 - \left( {\dfrac{{b - c}}{{b + c}}} \right)}}$

$\Rightarrow \tan \left( {\dfrac{\alpha }{2} + \dfrac{\beta }{2}} \right) = \dfrac{{\dfrac{{ - 2a}}{{b + c}}}}{{\dfrac{{b + c - b + c}}{{b + c}}}}$

$\Rightarrow \tan \left( {\dfrac{\alpha }{2} + \dfrac{\beta }{2}} \right) = \dfrac{{ - 2a}}{{2c}}$

$\Rightarrow \tan \left( {\dfrac{{\alpha + \beta }}{2}} \right) = \dfrac{{ - a}}{c}......\left( i \right)$

We know that $\tan 2x = \dfrac{{2\tan x}}{{1 - {{\tan }^2}x}}$. Thus, we have

$\Rightarrow \tan 2\left( {\dfrac{{\alpha + \beta }}{2}} \right) = \dfrac{{2\tan \left( {\dfrac{{\alpha + \beta }}{2}} \right)}}{{1 - {{\tan }^2}\left( {\dfrac{{\alpha + \beta }}{2}} \right)}}$

On substituting the values, we get

$\Rightarrow \tan \left( {\alpha + \beta } \right) = \dfrac{{2\left( {\dfrac{{ - a}}{c}} \right)}}{{1 - {{\left( {\dfrac{{ - a}}{c}} \right)}^2}}}$

$\Rightarrow \tan \left( {\alpha + \beta } \right) = \dfrac{{\dfrac{{ - 2a}}{c}}}{{\dfrac{{{c^2} - {a^2}}}{{{c^2}}}}}$

$\Rightarrow \tan \left( {\alpha + \beta } \right) = \dfrac{{ - 2a}}{c} \times \dfrac{{{c^2}}}{{{c^2} - {a^2}}}$

$\Rightarrow \tan \left( {\alpha + \beta } \right) = \dfrac{{ - 2ac}}{{{c^2} - {a^2}}}$

$\Rightarrow \tan \left( {\alpha + \beta } \right) = \dfrac{{2ac}}{{{a^2} - {c^2}}}$

Hence proved

Example 12: Show that $2{\sin ^2}\beta + 4\cos \left( {\alpha + \beta } \right)\sin \alpha \sin \beta + \cos 2\left( {\alpha + \beta } \right) = \cos 2\alpha$

Ans: We have LHS, $2{\sin ^2}\beta + 4\cos \left( {\alpha + \beta } \right)\sin \alpha \sin \beta + \cos 2\left( {\alpha + \beta } \right)$

$2{\sin ^2}\beta + 4\cos \left( {\alpha + \beta } \right)\sin \alpha \sin \beta + \cos \left( {2\alpha + 2\beta } \right)$

We know that $\cos \left( {A + B} \right) = \cos A\cos B - \sin A\sin B$. Therefore, we get

$\Rightarrow 2{\sin ^2}\beta + 4\left( {\cos \alpha \cos \beta - \sin \alpha \sin \beta } \right)\sin \alpha \sin \beta + \left( {\cos 2\alpha \cos 2\beta - \sin 2\alpha \sin 2\beta } \right)$

$\Rightarrow 2{\sin ^2}\beta + 4\sin \alpha \cos \alpha \sin \beta \cos \beta - 4{\sin ^2}\alpha {\sin ^2}\beta + \cos 2\alpha \cos 2\beta - \sin 2\alpha \sin 2\beta$

We know that $2\sin x\cos x = \sin 2x$. Therefore, we get

$\Rightarrow 2{\sin ^2}\beta + \sin 2\alpha \sin 2\beta - 4{\sin ^2}\alpha {\sin ^2}\beta + \cos 2\alpha \cos 2\beta - \sin 2\alpha \sin 2\beta$

$\Rightarrow 2{\sin ^2}\beta - 4{\sin ^2}\alpha {\sin ^2}\beta + \cos 2\alpha \cos 2\beta$

We can also the above written expression as,

$\Rightarrow 2{\sin ^2}\beta - \left( {2{{\sin }^2}\alpha } \right)\left( {2{{\sin }^2}\beta } \right) + \cos 2\alpha \cos 2\beta$

We know that $2{\sin ^2}A = 1 - \cos 2A$. Therefore, we get

$\Rightarrow \left( {1 - \cos 2\beta } \right) - \left( {1 - \cos 2\alpha } \right)\left( {1 - \cos 2\beta } \right) + \cos 2\alpha \cos 2\beta$

$\Rightarrow 1 - \cos 2\beta - \left( {1 - \cos 2\beta - \cos 2\alpha + \cos 2\beta \cos 2\alpha } \right) + \cos 2\alpha \cos 2\beta$

$\Rightarrow 1 - \cos 2\beta - 1 + \cos 2\beta + \cos 2\alpha - \cos 2\beta \cos 2\alpha + \cos 2\alpha \cos 2\beta$

$\Rightarrow \cos 2\alpha$

Hence proved

Example 13: If an angle $\theta$ is divided into two parts such that the tangent of one part is $k$ times the tangent of other, and $\phi$ is their difference, then show that $\sin \theta = \dfrac{{k + 1}}{{k - 1}}\sin \phi$.

Ans: Let $\theta = \alpha + \beta$. Then $\tan \alpha = k\tan \beta$.

$\Rightarrow \dfrac{{\tan \alpha }}{{\tan \beta }} = \dfrac{k}{1}$

Now, we will apply componendo and dividend

$\Rightarrow \dfrac{{\tan \alpha + \tan \beta }}{{\tan \alpha - \tan \beta }} = \dfrac{{k + 1}}{{k - 1}}$

Now, we will write the above written expression in terms of $\sin e$ and $\cos ine$

$\Rightarrow \dfrac{{\dfrac{{\sin \alpha }}{{\cos \alpha }} + \dfrac{{\sin \beta }}{{\cos \beta }}}}{{\dfrac{{\sin \alpha }}{{\cos \alpha }} - \dfrac{{\sin \beta }}{{\cos \beta }}}} = \dfrac{{k + 1}}{{k - 1}}$

$\Rightarrow \dfrac{{\dfrac{{\sin \alpha \cos \beta + \cos \alpha \sin \beta }}{{\cos \alpha \cos \beta }}}}{{\dfrac{{\sin \alpha \cos \beta - \cos \alpha \sin \beta }}{{\cos \alpha \cos \beta }}}} = \dfrac{{k + 1}}{{k - 1}}$

$\Rightarrow \dfrac{{\sin \alpha \cos \beta + \cos \alpha \sin \beta }}{{\sin \alpha \cos \beta - \cos \alpha \sin \beta }} = \dfrac{{k + 1}}{{k - 1}}$

We know that $\sin \left( {A + B} \right) = \sin A\cos B + \cos A\sin B$ and $\sin \left( {A - B} \right) = \sin A\cos B - \cos A\sin B$. Therefore, we get

$\Rightarrow \dfrac{{\sin \left( {\alpha + \beta } \right)}}{{\sin \left( {\alpha - \beta } \right)}} = \dfrac{{k + 1}}{{k - 1}}$

Given that, $\alpha - \beta = \phi$ and $\alpha + \beta = \theta$. Therefore, we get

$\Rightarrow \dfrac{{\sin \theta }}{{\sin \phi }} = \dfrac{{k + 1}}{{k - 1}}$

$\Rightarrow \sin \theta = \dfrac{{k + 1}}{{k - 1}}\sin \phi$

Hence proved

Example 14: Solve $\sqrt 3 \cos \theta + \sin \theta = \sqrt 2$.

Ans: We have equation, $\sqrt 3 \cos \theta + \sin \theta = \sqrt 2$

Divide the equation by $2$

$\Rightarrow \dfrac{{\sqrt 3 }}{2}\cos \theta + \dfrac{1}{2}\sin \theta = \dfrac{{\sqrt 2 }}{2}$

$\Rightarrow \dfrac{{\sqrt 3 }}{2}\cos \theta + \dfrac{1}{2}\sin \theta = \dfrac{1}{{\sqrt 2 }}$

We know that $\cos \dfrac{\pi }{6} = \dfrac{{\sqrt 3 }}{2}$, $\cos \dfrac{\pi }{4} = \dfrac{1}{{\sqrt 2 }}$and $\sin \dfrac{\pi }{6} = \dfrac{1}{2}$. So, we can written above-written equation as

$\Rightarrow \cos \dfrac{\pi }{6}\cos \theta + \sin \dfrac{\pi }{6}\sin \theta = \cos \dfrac{\pi }{4}$

$\Leftrightarrow \cos \theta \cos \dfrac{\pi }{6} + \sin \theta \sin \dfrac{\pi }{6} = \cos \dfrac{\pi }{4}$

We know that $\cos \left( {x - y} \right) = \cos x\cos y + \sin x\sin y$. Therefore, we get

$\Rightarrow \cos \left( {\theta - \dfrac{\pi }{6}} \right) = \cos \dfrac{\pi }{4}$

We know that when $\cos \theta = \cos \alpha$, then $\theta = 2n\pi \pm \alpha$, where $n \in Z$.

$\Rightarrow \theta - \dfrac{\pi }{6} = 2n\pi \pm \dfrac{\pi }{4}$

$\Rightarrow \theta = 2n\pi \pm \dfrac{\pi }{4} + \dfrac{\pi }{6}$

Hence, the solutions are $\theta = 2n\pi + \dfrac{\pi }{4} + \dfrac{\pi }{6}$ and $\theta = 2n\pi - \dfrac{\pi }{4} + \dfrac{\pi }{6}$

i.e., $\theta = 2n\pi + \dfrac{{5\pi }}{{12}}$ and $\theta = 2n\pi - \dfrac{\pi }{{12}}$

Objective Type Questions

Choose the correct answer from the given four options against each of the examples $15$ to $19$.

Example 15: If $\tan \theta = \dfrac{{ - 4}}{3}$, then $\sin \theta$ is

a) $\dfrac{{ - 4}}{5}$ but not $\dfrac{4}{5}$

b) $\dfrac{{ - 4}}{5}$ or $\dfrac{4}{5}$

c) $\dfrac{4}{5}$ but not $\dfrac{{ - 4}}{5}$

d) None of these

Ans: The correct answer is option (b) $\dfrac{{ - 4}}{5}$ or $\dfrac{4}{5}$

Given that, $\tan \theta = \dfrac{{ - 4}}{3} = \dfrac{P}{B}$.

By Pythagoras theorem, we have

$\Rightarrow {H^2} = {P^2} + {B^2}$

$\Rightarrow {H^2} = {4^2} + {3^2}$

(Here, we have taken positive value of perpendicular because length can’t be negative)

$\Rightarrow {H^2} = 16 + 9$

$\Rightarrow {H^2} = 25$

$\Rightarrow H = 5$

Since $\tan \theta = \dfrac{{ - 4}}{3}$ is negative, $\theta$ lies either in second quadrant or in fourth quadrant.

We know that $\sin \theta = \dfrac{P}{H}$. Therefore, we get

If $\theta$ lies in second quadrant, $\sin \theta = \dfrac{4}{5}$ and if $\theta$ lies in fourth quadrant, $\sin \theta = - \dfrac{4}{5}$.

Hence, the required answer is (b) $\dfrac{{ - 4}}{5}$ or $\dfrac{4}{5}$

Example 16: If $\sin \theta$ and $\cos \theta$ are the roots of the equation $a{x^2} - bx + c = 0$, then $a$, $b$ and $c$ satisfy the relation.

a) ${a^2} + {b^2} + 2ac = 0$

b) ${a^2} - {b^2} + 2ac = 0$

c) ${a^2} + {c^2} + 2ab = 0$

d) ${a^2} - {b^2} - 2ac = 0$

Ans: The correct answer is option (b) ${a^2} - {b^2} + 2ac = 0$

Given that, $\sin \theta$ and $\cos \theta$ are the roots of the equation $a{x^2} - bx + c = 0$.

We know that if the roots of the quadratic equation $a{x^2} + bx + c = 0$ are $\alpha$ and $\beta$. Then we have, $\alpha + \beta = - \dfrac{b}{a}$ and $\alpha \beta = \dfrac{c}{a}$. Therefore, we get

$\Rightarrow \sin \theta + \cos \theta = \dfrac{b}{a}...\left( i \right)$ and $\sin \theta \cos \theta = \dfrac{c}{a}.....\left( {ii} \right)$

On squaring both the sides in equation $\left( i \right)$, we get

$\Rightarrow {\sin ^2}\theta + {\cos ^2}\theta + 2\sin \theta \cos \theta = \dfrac{{{b^2}}}{{{a^2}}}$

We have, $\sin \theta \cos \theta = \dfrac{c}{a}$ and we know that ${\sin ^2}\theta + {\cos ^2}\theta = 1$. Therefore, we get

$\Rightarrow 1 + \dfrac{{2c}}{a} = \dfrac{{{b^2}}}{{{a^2}}}$

$\Rightarrow \dfrac{{a + 2c}}{a} = \dfrac{{{b^2}}}{{{a^2}}}$

$\Rightarrow \dfrac{{a + 2c}}{1} = \dfrac{{{b^2}}}{a}$

On cross multiplication, we get

$\Rightarrow {a^2} + 2ac = {b^2}$

$\Rightarrow {a^2} - {b^2} + 2ac = 0$

Example 17: The greatest value of $\sin x\cos x$ is

a) $1$

b) $2$

c) $\sqrt 2$

d) $\dfrac{1}{2}$

Ans: The correct answer is option (d) $\dfrac{1}{2}$

We have, $\sin x\cos x$

Multiply and divide the expression by $2$

$\Rightarrow \dfrac{1}{2} \times 2\sin x\cos x$

We know that $2\sin x\cos x = \sin 2x$. Therefore, we get

$\Rightarrow \dfrac{1}{2} \times \sin 2x$

We know that,

$\Rightarrow - 1 \leqslant \sin 2x \leqslant 1$

Divide the expression by $2$

$\Rightarrow - \dfrac{1}{2} \leqslant \dfrac{{\sin 2x}}{2} \leqslant \dfrac{1}{2}$

Hence, the greatest is $\dfrac{1}{2}$.

Example 18: The value of $\sin 20^\circ \sin 40^\circ \sin 60^\circ \sin 80^\circ$ is

a) $\dfrac{{ - 3}}{{16}}$

b) $\dfrac{5}{{16}}$

c) $\dfrac{3}{{16}}$

d) $\dfrac{1}{{16}}$

Ans: The correct answer is option (c) $\dfrac{3}{{16}}$

$\Rightarrow \sin 60^\circ \sin 20^\circ \sin 40^\circ \sin 80^\circ$

We know that $\sin 60^\circ = \dfrac{{\sqrt 3 }}{2}$. Therefore, we get

$\Rightarrow \dfrac{{\sqrt 3 }}{2}\sin 20^\circ \left( {\sin 40^\circ \sin 80^\circ } \right)$

Multiply and divide the expression by $2$.

$\Rightarrow \dfrac{{\sqrt 3 }}{2} \times \dfrac{1}{2}\sin 20^\circ \left( {2\sin 80^\circ \sin 40^\circ } \right)$

We know that $2\sin A\sin B = \cos \left( {A - B} \right) - \cos \left( {A + B} \right)$. Therefore, we get

$\Rightarrow \dfrac{{\sqrt 3 }}{4}\sin 20^\circ \left( {\cos 40^\circ - \cos 120^\circ } \right)$

$\Rightarrow \dfrac{{\sqrt 3 }}{4}\left[ {\sin 20^\circ \cos 40^\circ - \sin 20^\circ \cos 120^\circ } \right]$

We know that $\cos 120^\circ = - \dfrac{1}{2}$. Therefore, we get

$\Rightarrow \dfrac{{\sqrt 3 }}{4}\left[ {\sin 20^\circ \cos 40^\circ - \sin 20^\circ \left( { - \dfrac{1}{2}} \right)} \right]$

Multiply and divide the expression by $2$

$\Rightarrow \dfrac{{\sqrt 3 }}{{4 \times 2}}\left[ {2\sin 20^\circ \cos 40^\circ - 2\sin 20^\circ \left( { - \dfrac{1}{2}} \right)} \right]$

$\Rightarrow \dfrac{{\sqrt 3 }}{8}\left[ {2\sin 20^\circ \cos 40^\circ + \sin 20^\circ } \right]$

We know that $2\sin A\cos B = \sin \left( {A + B} \right) + \sin \left( {A - B} \right)$. Therefore, we get

$\Rightarrow \dfrac{{\sqrt 3 }}{8}\left[ {\sin 60^\circ + \sin \left( { - 20^\circ } \right) + \sin 20^\circ } \right]$

We know that $\sin \left( { - \theta } \right) = - \sin \theta$. Therefore, we get

$\Rightarrow \dfrac{{\sqrt 3 }}{8}\left( {\dfrac{{\sqrt 3 }}{2}} \right)$

$\Rightarrow \dfrac{3}{{16}}$

Example 19: The value of $\cos \dfrac{\pi }{5}\cos \dfrac{{2\pi }}{5}\cos \dfrac{{4\pi }}{5}\cos \dfrac{{8\pi }}{5}$ is

a) $\dfrac{1}{{16}}$

b) $0$

c) $\dfrac{{ - 1}}{8}$

d) $\dfrac{{ - 1}}{{16}}$

Ans: The correct answer is option (d) $\dfrac{{ - 1}}{{16}}$

We have, $\cos \dfrac{\pi }{5}\cos \dfrac{{2\pi }}{5}\cos \dfrac{{4\pi }}{5}\cos \dfrac{{8\pi }}{5}$

Multiply the above-written expression by $2\sin \dfrac{\pi }{5}$.

$\Rightarrow \dfrac{1}{{2\sin \dfrac{\pi }{5}}}2\sin \dfrac{\pi }{5}\cos \dfrac{\pi }{5}\cos \dfrac{{2\pi }}{5}\cos \dfrac{{4\pi }}{5}\cos \dfrac{{8\pi }}{5}$

We know that $2\sin x\cos x = \sin 2x$. Therefore, we get

$\Rightarrow \dfrac{1}{{2\sin \dfrac{\pi }{5}}}\sin \dfrac{{2\pi }}{5}\cos \dfrac{{2\pi }}{5}\cos \dfrac{{4\pi }}{5}\cos \dfrac{{8\pi }}{5}$

Multiply and divide the expression by $2$

$\Rightarrow \dfrac{1}{{2 \times 2\sin \dfrac{\pi }{5}}}2\sin \dfrac{{2\pi }}{5}\cos \dfrac{{2\pi }}{5}\cos \dfrac{{4\pi }}{5}\cos \dfrac{{8\pi }}{5}$

We know that $2\sin x\cos x = \sin 2x$. Therefore, we get

$\Rightarrow \dfrac{1}{{4\sin \dfrac{\pi }{5}}}\sin \dfrac{{4\pi }}{5}\cos \dfrac{{4\pi }}{5}\cos \dfrac{{8\pi }}{5}$

Multiply and divide the expression by $2$

$\Rightarrow \dfrac{1}{{2 \times 4\sin \dfrac{\pi }{5}}}2\sin \dfrac{{4\pi }}{5}\cos \dfrac{{4\pi }}{5}\cos \dfrac{{8\pi }}{5}$

$\Rightarrow \dfrac{1}{{8\sin \dfrac{\pi }{5}}}\sin \dfrac{{8\pi }}{5}\cos \dfrac{{8\pi }}{5}$

Multiply and divide the expression by $2$

$\Rightarrow \dfrac{1}{{2 \times 8\sin \dfrac{\pi }{5}}}2\sin \dfrac{{8\pi }}{5}\cos \dfrac{{8\pi }}{5}$

$\Rightarrow \dfrac{1}{{16\sin \dfrac{\pi }{5}}}\sin \dfrac{{16\pi }}{5}$

$\Rightarrow \dfrac{{\sin \left( {3\pi + \dfrac{\pi }{5}} \right)}}{{16\sin \dfrac{\pi }{5}}}$

$\Rightarrow \dfrac{{ - \sin \dfrac{\pi }{5}}}{{16\sin \dfrac{\pi }{5}}}$

$\Rightarrow \dfrac{{ - 1}}{{16}}$

Fill in the blank:

Example 20: If $3\tan \left( {\theta - 15^\circ } \right) = \tan \left( {\theta + 15^\circ } \right)$, $0^\circ < \theta < 90^\circ$, then $\theta$ = _____.

Ans: Given that, $3\tan \left( {\theta - 15^\circ } \right) = \tan \left( {\theta + 15^\circ } \right)$

$\Rightarrow \dfrac{{\tan \left( {\theta + 15^\circ } \right)}}{{\tan \left( {\theta - 15^\circ } \right)}} = \dfrac{3}{1}$

Now, we will apply componendo and dividendo rule. Therefore, we get

$\Rightarrow \dfrac{{\tan \left( {\theta + 15^\circ } \right) + \tan \left( {\theta - 15^\circ } \right)}}{{\tan \left( {\theta + 15^\circ } \right) - \tan \left( {\theta - 15^\circ } \right)}} = \dfrac{{3 + 1}}{{3 - 1}}$

$\Rightarrow \dfrac{{\dfrac{{\sin \left( {\theta + 15^\circ } \right)}}{{\cos \left( {\theta + 15^\circ } \right)}} + \dfrac{{\sin \left( {\theta - 15^\circ } \right)}}{{\cos \left( {\theta - 15^\circ } \right)}}}}{{\dfrac{{\sin \left( {\theta + 15^\circ } \right)}}{{\cos \left( {\theta + 15^\circ } \right)}} - \dfrac{{\sin \left( {\theta - 15^\circ } \right)}}{{\cos \left( {\theta - 15^\circ } \right)}}}} = \dfrac{4}{2}$

$\Rightarrow \dfrac{{\dfrac{{\sin \left( {\theta + 15^\circ } \right)\cos \left( {\theta - 15^\circ } \right) + \sin \left( {\theta - 15^\circ } \right)\cos \left( {\theta + 15^\circ } \right)}}{{\cos \left( {\theta + 15^\circ } \right)\cos \left( {\theta - 15^\circ } \right)}}}}{{\dfrac{{\sin \left( {\theta + 15^\circ } \right)\cos \left( {\theta - 15^\circ } \right) - \sin \left( {\theta - 15^\circ } \right)\cos \left( {\theta + 15^\circ } \right)}}{{\cos \left( {\theta + 15^\circ } \right)\cos \left( {\theta - 15^\circ } \right)}}}} = 2$

$\Rightarrow \dfrac{{\sin \left( {\theta + 15^\circ } \right)\cos \left( {\theta - 15^\circ } \right) + \sin \left( {\theta - 15^\circ } \right)\cos \left( {\theta + 15^\circ } \right)}}{{\sin \left( {\theta + 15^\circ } \right)\cos \left( {\theta - 15^\circ } \right) - \sin \left( {\theta - 15^\circ } \right)\cos \left( {\theta + 15^\circ } \right)}} = 2$

We know that $\sin \left( {A + B} \right) = \sin A\cos B + \cos A\sin B$ and $\sin \left( {A - B} \right) = \sin A\cos B - \cos A\sin B$. Therefore, we get

$\Rightarrow \dfrac{{\sin \left( {\theta + 15^\circ + \theta - 15^\circ } \right)}}{{\sin \left( {\theta + 15^\circ - \left( {\theta - 15^\circ } \right)} \right)}} = 2$

$\Rightarrow \dfrac{{\sin 2\theta }}{{\sin \left( {\theta + 15^\circ - \theta + 15^\circ } \right)}} = 2$

$\Rightarrow \dfrac{{\sin 2\theta }}{{\sin 30^\circ }} = 2$

We know that $\sin 30^\circ = \dfrac{1}{2}$. Therefore, we get

$\Rightarrow \dfrac{{\sin 2\theta }}{{\dfrac{1}{2}}} = 2$

$\Rightarrow \dfrac{{2\sin 2\theta }}{1} = 2$

$\Rightarrow \sin 2\theta = 1$

Given that, $0^\circ < \theta < 90^\circ$ i.e., $0 < \theta < \dfrac{\pi }{2}$

$\therefore 0 < 2\theta < \pi$

$\Rightarrow \sin 2\theta = \sin \dfrac{\pi }{2}$

$\Rightarrow 2\theta = \dfrac{\pi }{2}$

$\Rightarrow \theta = \dfrac{\pi }{4}$

Or

$\Rightarrow \theta = 45^\circ$

State whether the following statement is true or false. Justify your answer.

Example 21: “The inequality ${2^{\sin \theta }} + {2^{\cos \theta }} \geqslant {2^{1 - \dfrac{1}{{\sqrt 2 }}}}$ holds for all real values of $\theta$ ”.

Ans: The given statement is true.

Since, ${2^{\sin \theta }}$ and ${2^{\cos \theta }}$ are positive real numbers, so arithmetic mean of these two numbers is greater or equal to their geometric mean.

$\Rightarrow \dfrac{{{2^{\sin \theta }} + {2^{\cos \theta }}}}{2} \geqslant \sqrt {{2^{\sin \theta }} \times {2^{\cos \theta }}}$

$\Rightarrow {2^{\sin \theta }} + {2^{\cos \theta }} \geqslant 2\sqrt {{2^{\sin \theta + \cos \theta }}}$

$\Rightarrow {2^{\sin \theta }} + {2^{\cos \theta }} \geqslant {2.2^{\dfrac{{\sin \theta + \cos \theta }}{2}}}......\left( i \right)$

We have, $\sin \theta + \cos \theta$.

Divide and multiply the expression by $\sqrt 2$

$\Rightarrow \sqrt 2 \left( {\sin \theta \times \dfrac{1}{{\sqrt 2 }} + \cos \theta \times \dfrac{1}{{\sqrt 2 }}} \right)$

We know that $\sin \dfrac{\pi }{4} = \dfrac{1}{{\sqrt 2 }}$ and $\cos \dfrac{\pi }{4} = \dfrac{1}{{\sqrt 2 }}$. Therefore, we get

$\Rightarrow \sqrt 2 \left( {\sin \theta \cos \dfrac{\pi }{4} + \cos \theta \sin \dfrac{\pi }{4}} \right)$

We know that $\sin \left( {x + y} \right) = \sin x\cos y + \cos x\sin y$. Therefore, we get

$\Rightarrow \sqrt 2 \sin \left( {\theta + \dfrac{\pi }{4}} \right)$

Since, $- 1 \leqslant \sin \left( {\theta + \dfrac{\pi }{4}} \right) \leqslant 1$

$\Rightarrow - \sqrt 2 \leqslant \sqrt 2 \sin \left( {\theta + \dfrac{\pi }{4}} \right) \leqslant \sqrt 2$

As we solved above, $\sin \theta + \cos \theta = \sqrt 2 \sin \left( {\theta + \dfrac{\pi }{4}} \right)$. Therefore, we get

$\Rightarrow - \sqrt 2 \leqslant \sin \theta + \cos \theta \leqslant \sqrt 2$

$\Rightarrow - \dfrac{{\sqrt 2 }}{2} \leqslant \dfrac{{\sin \theta + \cos \theta }}{2} \leqslant \dfrac{{\sqrt 2 }}{2}$

$\Rightarrow - \dfrac{1}{{\sqrt 2 }} \leqslant \dfrac{{\sin \theta + \cos \theta }}{2} \leqslant \dfrac{1}{{\sqrt 2 }}$

We have, ${2^{\sin \theta }} + {2^{\cos \theta }} \geqslant {2.2^{\dfrac{{\sin \theta + \cos \theta }}{2}}}......\left( i \right)$

$\Rightarrow {2^{\sin \theta }} + {2^{\cos \theta }} \geqslant {2.2^{\dfrac{{ - 1}}{{\sqrt 2 }}}}$

$\Rightarrow {2^{\sin \theta }} + {2^{\cos \theta }} \geqslant {2^{1 + \left( {\dfrac{{ - 1}}{{\sqrt 2 }}} \right)}}$

$\Rightarrow {2^{\sin \theta }} + {2^{\cos \theta }} \geqslant {2^{1 - \dfrac{1}{{\sqrt 2 }}}}$

State whether the following statement is true or false. Justify your answer.

Example 22: Match each item given under the column ${C_1}$ to its correct answer given under the column ${C_2}$.

${C_1}$                                                                            ${C_2}$

a) $\dfrac{{1 - \cos x}}{{\sin x}}$                                                                 (i) ${\cot ^2}\dfrac{x}{2}$

b) $\dfrac{{1 + \cos x}}{{1 - \cos x}}$                                                                 (ii) $\cot \dfrac{x}{2}$

c) $\dfrac{{1 + \cos x}}{{\sin x}}$                                                                  (iii) $\left| {\cos x + \sin x} \right|$

d) $\sqrt {1 + \sin 2x}$                                                              (iv) $\tan \dfrac{x}{2}$

Ans:

a) $\dfrac{{1 - \cos x}}{{\sin x}}$

We know that $1 - \cos 2A = 2{\sin ^2}A$ and $\sin 2A = 2\sin A\cos A$. Therefore, we get

$\Rightarrow \dfrac{{2{{\sin }^2}\dfrac{x}{2}}}{{2\sin \dfrac{x}{2}\cos \dfrac{x}{2}}}$

$\Rightarrow \dfrac{{\sin \dfrac{x}{2}}}{{\cos \dfrac{x}{2}}}$

$\Rightarrow \tan \dfrac{x}{2}$

Hence, $\left( a \right) \leftrightarrow \left( {iv} \right)$.

b) $\dfrac{{1 + \cos x}}{{1 - \cos x}}$

We know that $1 + \cos 2A = 2{\cos ^2}A$ and $1 - \cos 2A = 2{\sin ^2}A$. Therefore, we get

$\Rightarrow \dfrac{{2{{\cos }^2}\dfrac{x}{2}}}{{2{{\sin }^2}\dfrac{x}{2}}}$

$\Rightarrow {\cot ^2}\dfrac{x}{2}$

Hence, $\left( b \right) \leftrightarrow \left( i \right)$.

c) $\dfrac{{1 + \cos x}}{{\sin x}}$

We know that $1 + \cos 2A = 2{\cos ^2}A$ and $\sin 2A = 2\sin A\cos A$. Therefore, we get

$\Rightarrow \dfrac{{2{{\cos }^2}\dfrac{x}{2}}}{{2\sin \dfrac{x}{2}\cos \dfrac{x}{2}}}$

$\Rightarrow \dfrac{{\cos \dfrac{x}{2}}}{{\sin \dfrac{x}{2}}}$

$\Rightarrow \cot \dfrac{x}{2}$

Hence, $\left( c \right) \leftrightarrow \left( {ii} \right)$

d) $\sqrt {1 + \sin 2x}$

We know that ${\sin ^2}A + {\cos ^2}A = 1$ and $\sin 2A = 2\sin A\cos A$. Therefore, we get

$\Rightarrow \sqrt {{{\sin }^2}x + {{\cos }^2}x + 2\sin x\cos x}$

$\Rightarrow \sqrt {{{\left( {\sin x + \cos x} \right)}^2}}$

$\Rightarrow \left| {\left( {\sin x + \cos x} \right)} \right|$

$\Rightarrow \left| {\cos x + \sin x} \right|$

Hence, $\left( d \right) \leftrightarrow \left( {iii} \right)$.

Exercise 3.3

Short Answer Type

1. Prove that $\dfrac{{\tan A + \sec A - 1}}{{\tan A - \sec A + 1}} = \dfrac{{1 + \sin A}}{{\cos A}}$.

Ans: We need to prove that $\dfrac{{\tan A + \sec A - 1}}{{\tan A - \sec A + 1}} = \dfrac{{1 + \sin A}}{{\cos A}}$

L.H.S. = $\dfrac{{\tan A + \sec A - 1}}{{\tan A - \sec A + 1}}$

We know that ${\sec ^2}A - {\tan ^2}A = 1$. Therefore, we get

$\Rightarrow \dfrac{{\tan A + \sec A - \left( {{{\sec }^2}A - {{\tan }^2}A} \right)}}{{\tan A - \sec A + 1}}$

We know that ${a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)$. Therefore, we get

$\Rightarrow \dfrac{{\tan A + \sec A - \left[ {\left( {\sec A + \tan A} \right)\left( {\sec A - \tan A} \right)} \right]}}{{\tan A - \sec A + 1}}$

Take $\sec A + \tan A$ as a common term

$\Rightarrow \dfrac{{\left( {\sec A + \tan A} \right)\left[ {1 - \left( {\sec A - \tan A} \right)} \right]}}{{\tan A - \sec A + 1}}$

$\Rightarrow \dfrac{{\left( {\sec A + \tan A} \right)\left[ {1 - \sec A + \tan A} \right]}}{{\tan A - \sec A + 1}}$

On canceling common term, we get

$\Rightarrow \sec A + \tan A$

Now, we will convert above written expression in terms of $\sin e$ and $\cos$.

$\Rightarrow \dfrac{1}{{\cos A}} + \dfrac{{\sin A}}{{\cos A}}$

Take LCM

$\Rightarrow \dfrac{{1 + \sin A}}{{\cos A}}$

Hence proved.

1. If $\dfrac{{2\sin \alpha }}{{1 + \cos \alpha + \sin \alpha }} = y$, prove that $\dfrac{{1 - \cos \alpha + \sin \alpha }}{{1 + \sin \alpha }}$ is also equal to $y$.

Ans: Given, $y = \dfrac{{2\sin \alpha }}{{1 + \cos \alpha + \sin \alpha }}$

Let us multiply and divide the above written expression by $1 + \sin \alpha - \cos \alpha$.

$\Rightarrow \dfrac{{2\sin \alpha }}{{1 + \cos \alpha + \sin \alpha }} \times \dfrac{{1 + \sin \alpha - \cos \alpha }}{{1 + \sin \alpha - \cos \alpha }}$

$\Rightarrow \dfrac{{2\sin \alpha \left( {1 + \sin \alpha - \cos \alpha } \right)}}{{\left( {1 + \cos \alpha + \sin \alpha } \right)\left( {1 + \sin \alpha - \cos \alpha } \right)}}$

Now we will apply $\left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2}$ formula to simplify the denominator.

$\Rightarrow \dfrac{{2\sin \alpha \left( {1 + \sin \alpha - \cos \alpha } \right)}}{{{{\left( {1 + \sin \alpha } \right)}^2} - {{\cos }^2}\alpha }}$

On expansion using ${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$ , we get

$\Rightarrow \dfrac{{2\sin \alpha \left( {1 + \sin \alpha - \cos \alpha } \right)}}{{{1^2} + {{\sin }^2}\alpha + 2\sin \alpha - {{\cos }^2}\alpha }}$

$\Rightarrow \dfrac{{2\sin \alpha \left( {1 + \sin \alpha - \cos \alpha } \right)}}{{\left( {{1^2} - {{\cos }^2}\alpha } \right) + {{\sin }^2}\alpha + 2\sin \alpha }}$

We know that $1 - {\cos ^2}x = {\sin ^2}x$. Therefore, we get

$\Rightarrow \dfrac{{2\sin \alpha \left( {1 + \sin \alpha - \cos \alpha } \right)}}{{{{\sin }^2}\alpha + {{\sin }^2}\alpha + 2\sin \alpha }}$

$\Rightarrow \dfrac{{2\sin \alpha \left( {1 + \sin \alpha - \cos \alpha } \right)}}{{2{{\sin }^2}\alpha + 2\sin \alpha }}$

Take $2\sin \alpha$ as a common term

$\Rightarrow \dfrac{{2\sin \alpha \left( {1 + \sin \alpha - \cos \alpha } \right)}}{{2\sin \alpha \left( {\sin \alpha + 1} \right)}}$

On canceling common term, we get

$\Rightarrow \dfrac{{1 + \sin \alpha - \cos \alpha }}{{\sin \alpha + 1}} = y$

Hence proved.

1. If $m\sin \theta = n\sin \left( {\theta + 2\alpha } \right)$, then prove that $\tan \left( {\theta + \alpha } \right)\cot \alpha = \dfrac{{m + n}}{{m - n}}$.

Ans: Given, $m\sin \theta = n\sin \left( {\theta + 2\alpha } \right)$

We can also write it as,

$\Rightarrow \dfrac{{\sin \left( {\theta + 2\alpha } \right)}}{{\sin \theta }} = \dfrac{m}{n}$

Now, we will apply componendo and dividend rule on the above written expression.

$\Rightarrow \dfrac{{\sin \left( {\theta + 2\alpha } \right) + \sin \theta }}{{\sin \left( {\theta + 2\alpha } \right) - \sin \theta }} = \dfrac{{m + n}}{{m - n}}$

We know that $\sin A + \sin B = 2\sin \dfrac{{A + B}}{2}.\cos \dfrac{{A - B}}{2}$ and $\sin A - \sin B = 2\cos \dfrac{{A + B}}{2}.\sin \dfrac{{A - B}}{2}$. Therefore, we get

$\Rightarrow \dfrac{{2\sin \left( {\dfrac{{\theta + 2\alpha + \theta }}{2}} \right).\cos \left( {\dfrac{{\theta + 2\alpha - \theta }}{2}} \right)}}{{2\cos \left( {\dfrac{{\theta + 2\alpha + \theta }}{2}} \right).\sin \left( {\dfrac{{\theta + 2\alpha - \theta }}{2}} \right)}} = \dfrac{{m + n}}{{m - n}}$

$\Rightarrow \dfrac{{2\sin \left( {\dfrac{{2\theta + 2\alpha }}{2}} \right).\cos \left( {\dfrac{{2\alpha }}{2}} \right)}}{{2\cos \left( {\dfrac{{2\theta + 2\alpha }}{2}} \right).\sin \left( {\dfrac{{2\alpha }}{2}} \right)}} = \dfrac{{m + n}}{{m - n}}$

On simplification, we get

$\Rightarrow \dfrac{{2\sin \left( {\theta + \alpha } \right).\cos \left( \alpha \right)}}{{2\cos \left( {\theta + \alpha } \right).\sin \left( \alpha \right)}} = \dfrac{{m + n}}{{m - n}}$

Now we will write above written expression in terms of $\tan$ and $\cot$.

$\Rightarrow \tan \left( {\theta + \alpha } \right).\cot \alpha = \dfrac{{m + n}}{{m - n}}$

Hence proved.

1. If $\cos \left( {\alpha + \beta } \right) = \dfrac{4}{5}$ and $\sin \left( {\alpha - \beta } \right) = \dfrac{5}{{13}}$, where $\alpha$ lie between $0$ and $\dfrac{\pi }{4}$, find the value of $\tan 2\alpha$.

Ans: Given, $\cos \left( {\alpha + \beta } \right) = \dfrac{4}{5}$ and $\sin \left( {\alpha - \beta } \right) = \dfrac{5}{{13}}$

At first we will find the value $\tan \left( {\alpha + \beta } \right)$ and $\tan \left( {\alpha - \beta } \right)$.

For value of $\tan \left( {\alpha + \beta } \right)$, we have

$\Rightarrow \cos \left( {\alpha + \beta } \right) = \dfrac{4}{5} = \dfrac{B}{H}$

We know that ${H^2} = {P^2} + {B^2}$

$\Rightarrow {5^2} = {P^2} + {4^2}$

$\Rightarrow {P^2} = {5^2} - {4^2} = 25 - 16$

$\Rightarrow {P^2} = 9$

$\Rightarrow P = 3$

We know that $\tan x = \dfrac{P}{B}$. Therefore, we get

$\Rightarrow \tan \left( {\alpha + \beta } \right) = \dfrac{3}{4}$

For value of $\tan \left( {\alpha - \beta } \right)$, we have

$\Rightarrow \sin \left( {\alpha - \beta } \right) = \dfrac{5}{{13}} = \dfrac{P}{H}$

We know that ${H^2} = {P^2} + {B^2}$

$\Rightarrow {13^2} = {5^2} + {B^2}$

$\Rightarrow {B^2} = {13^2} - {5^2} = 169 - 25$

$\Rightarrow {B^2} = 144$

$\Rightarrow B = 12$

We know that $\tan x = \dfrac{P}{B}$. Therefore, we get

$\Rightarrow \tan \left( {\alpha - \beta } \right) = \dfrac{5}{{12}}$

As here we need to find value of $\tan 2\alpha$, we can write it as

$\Rightarrow \tan 2\alpha = \left[ {\alpha + \beta + \alpha - \beta } \right]$

$\Rightarrow \tan 2\alpha = \left[ {\left( {\alpha + \beta } \right) + \left( {\alpha - \beta } \right)} \right]$

We know that $\tan \left( {x + y} \right) = \dfrac{{\tan x + \tan y}}{{1 - \tan x\tan y}}$. Therefore, we get

$\Rightarrow \tan 2\alpha = \dfrac{{\tan \left( {\alpha + \beta } \right) + \tan \left( {\alpha - \beta } \right)}}{{1 - \tan \left( {\alpha + \beta } \right)\tan \left( {\alpha - \beta } \right)}}$

$\Rightarrow \tan 2\alpha = \dfrac{{\dfrac{3}{4} + \dfrac{5}{{12}}}}{{1 - \dfrac{3}{4} \times \dfrac{5}{{12}}}}$

$\Rightarrow \tan 2\alpha = \dfrac{{\dfrac{{9 + 5}}{{12}}}}{{\dfrac{{48 - 15}}{{48}}}}$

$\Rightarrow \tan 2\alpha = \dfrac{{14}}{{12}} \times \dfrac{{48}}{{33}}$

$\Rightarrow \tan 2\alpha = \dfrac{{56}}{{33}}$

This is our required answer.

1. If $\tan x = \dfrac{b}{a}$, then find the value of $\sqrt {\dfrac{{a + b}}{{a - b}}} + \sqrt {\dfrac{{a - b}}{{a + b}}}$.

Ans: Given, $\tan x = \dfrac{b}{a}$ and we need to find the value of $\sqrt {\dfrac{{a + b}}{{a - b}}} + \sqrt {\dfrac{{a - b}}{{a + b}}}$.

Let us first take LCM of the above written expression

$\Rightarrow \dfrac{{\sqrt {a + b} \times \sqrt {a + b} + \sqrt {a - b} \times \sqrt {a - b} }}{{\sqrt {a - b} \times \sqrt {a + b} }}$

On multiplication of terms, we get

$\Rightarrow \dfrac{{{{\left( {\sqrt {a + b} } \right)}^2} + {{\left( {\sqrt {a - b} } \right)}^2}}}{{\sqrt {\left( {a - b} \right)\left( {a + b} \right)} }}$

As we know $\left( {a - b} \right)\left( {a + b} \right) = {a^2} + {b^2}$. Therefore, we get

$\Rightarrow \dfrac{{a + b + a - b}}{{\sqrt {{a^2} - {b^2}} }}$

$\Rightarrow \dfrac{{2a}}{{a\sqrt {1 - \dfrac{{{b^2}}}{{{a^2}}}} }}$

We are given that $\tan x = \dfrac{b}{a}$. Therefore, we get

$\Rightarrow \dfrac{2}{{\sqrt {1 - {{\tan }^2}x} }}$

We can also write it as,

$\Rightarrow \dfrac{2}{{\sqrt {1 - \dfrac{{{{\sin }^2}x}}{{{{\cos }^2}x}}} }}$

On taking LCM, we get

$\Rightarrow \dfrac{2}{{\sqrt {\dfrac{{{{\cos }^2}x - {{\sin }^2}x}}{{{{\cos }^2}x}}} }}$

We know that ${\cos ^2}x - {\sin ^2}x = \cos 2x$. Therefore, we get

$\Rightarrow \dfrac{2}{{\dfrac{{\sqrt {\cos 2x} }}{{\cos x}}}}$

$\Rightarrow \dfrac{{2\cos x}}{{\sqrt {\cos 2x} }}$

Hence, $\sqrt {\dfrac{{a + b}}{{a - b}}} + \sqrt {\dfrac{{a - b}}{{a + b}}} = \dfrac{{2\cos x}}{{\sqrt {\cos 2x} }}$.

1. Prove that $\cos \theta \cos \dfrac{\theta }{2} - \cos 3\theta \cos \dfrac{{9\theta }}{2} = \sin 4\theta \sin \left( {\dfrac{{7\theta }}{2}} \right)$.

Ans: We need to prove that $\cos \theta \cos \dfrac{\theta }{2} - \cos 3\theta \cos \dfrac{{9\theta }}{2} = \sin 4\theta \sin \left( {\dfrac{{7\theta }}{2}} \right)$

Let us start to solve L.H.S. = $\cos \theta \cos \dfrac{\theta }{2} - \cos 3\theta \cos \dfrac{{9\theta }}{2}$

On multiplication and division of the above written expression by $2$, we get

$\Rightarrow \dfrac{1}{2}\left[ {2\cos \theta \cos \dfrac{\theta }{2}} \right] - \dfrac{1}{2}\left[ {2\cos 3\theta \cos \dfrac{{9\theta }}{2}} \right]$

Now, we will apply $\cos \left( {A + B} \right) + \cos \left( {A - B} \right) = 2\cos A\cos B$ formula.

$\Rightarrow \dfrac{1}{2}\left[ {\cos \left( {\theta + \dfrac{\theta }{2}} \right) + \cos \left( {\theta - \dfrac{\theta }{2}} \right)} \right] - \dfrac{1}{2}\left[ {\cos \left( {3\theta + \dfrac{{9\theta }}{2}} \right) + \cos \left( {3\theta - \dfrac{{9\theta }}{2}} \right)} \right]$

On simplification, we get

$\Rightarrow \dfrac{1}{2}\left[ {\cos \dfrac{{3\theta }}{2} + \cos \dfrac{\theta }{2}} \right] - \dfrac{1}{2}\left[ {\cos \dfrac{{15\theta }}{2} + \cos \left( { - \dfrac{{3\theta }}{2}} \right)} \right]$

$\Rightarrow \dfrac{1}{2}\left[ {\cos \dfrac{{3\theta }}{2} + \cos \dfrac{\theta }{2} - \cos \dfrac{{15\theta }}{2} - \cos \left( { - \dfrac{{3\theta }}{2}} \right)} \right]$

As we know $\cos \left( { - \theta } \right) = \cos \theta$. Therefore, we get

$\Rightarrow \dfrac{1}{2}\left[ {\cos \dfrac{{3\theta }}{2} + \cos \dfrac{\theta }{2} - \cos \dfrac{{15\theta }}{2} - \cos \dfrac{{3\theta }}{2}} \right]$

$\Rightarrow \dfrac{1}{2}\left[ {\cos \dfrac{\theta }{2} - \cos \dfrac{{15\theta }}{2}} \right]$

We know that $\cos C - \cos D = - 2\sin \left( {\dfrac{{C + D}}{2}} \right)\sin \left( {\dfrac{{C - D}}{2}} \right)$. Therefore, we get

$\Rightarrow \dfrac{1}{2}\left[ { - 2\sin \left( {\dfrac{{\dfrac{\theta }{2} + \dfrac{{15\theta }}{2}}}{2}} \right)\sin \left( {\dfrac{{\dfrac{\theta }{2} - \dfrac{{15\theta }}{2}}}{2}} \right)} \right]$

$\Rightarrow - \sin \left( {\dfrac{{\dfrac{{16\theta }}{2}}}{2}} \right)\sin \left( {\dfrac{{\dfrac{{ - 14\theta }}{2}}}{2}} \right)$

$\Rightarrow - \sin \left( {4\theta } \right)\sin \left( {\dfrac{{ - 7\theta }}{2}} \right)$

As we know $\sin \left( { - \theta } \right) = - \sin \theta$. Therefore, we get

$\Rightarrow \sin \left( {4\theta } \right)\sin \left( {\dfrac{{7\theta }}{2}} \right)$

Hence proved.

1. If $a\cos \theta + b\sin \theta = m$ and $a\sin \theta - b\cos \theta = n$, then show that ${a^2} + {b^2} = {m^2} + {n^2}$.

Ans: Given, $a\cos \theta + b\sin \theta = m$ and $a\sin \theta - b\cos \theta = n$

$\Rightarrow a\cos \theta + b\sin \theta = m$

On squaring both the side, we get

$\Rightarrow {\left( {a\cos \theta + b\sin \theta } \right)^2} = {m^2}$

$\Rightarrow {a^2}{\cos ^2}\theta + {b^2}{\sin ^2}\theta + 2ab\sin \theta \cos \theta = {m^2}.....\left( i \right)$

$\Rightarrow a\sin \theta - b\cos \theta = n$

On squaring both the side, we get

$\Rightarrow {\left( {a\sin \theta - b\cos \theta } \right)^2} = {n^2}$

$\Rightarrow {a^2}{\sin ^2}\theta + {b^2}{\cos ^2}\theta - 2ab\sin \theta \cos \theta = {n^2}.....\left( {ii} \right)$

Add equation $\left( i \right)$ and $\left( {ii} \right)$.

$\Rightarrow {m^2} + {n^2} = {a^2}{\cos ^2}\theta + {b^2}{\sin ^2}\theta + 2ab\sin \theta \cos \theta + {a^2}{\sin ^2}\theta + {b^2}{\cos ^2}\theta - 2ab\sin \theta \cos \theta$

$\Rightarrow {m^2} + {n^2} = {a^2}\left( {{{\cos }^2}\theta + {{\sin }^2}\theta } \right) + {b^2}\left( {{{\sin }^2}\theta + {{\cos }^2}\theta } \right)$

As we know ${\sin ^2}\theta + {\cos ^2}\theta = 1$. Therefore, we get

$\Rightarrow {m^2} + {n^2} = {a^2}\left( 1 \right) + {b^2}\left( 1 \right)$

$\Rightarrow {m^2} + {n^2} = {a^2} + {b^2}$

Hence proved.

1. Find the value of $\tan 20^\circ 30'$.

Ans: We need to find the value of $\tan 20^\circ 30'$.

Let $22^\circ 33' = \dfrac{\theta }{2}$

$\Rightarrow \theta = 45^\circ$

$\Rightarrow \tan 20^\circ 30' = \tan \dfrac{\theta }{2}$

We know that $\tan x = \dfrac{{\sin x}}{{\cos x}}$. Therefore, we get

$\Rightarrow \tan 20^\circ 30' = \dfrac{{\sin \dfrac{\theta }{2}}}{{\cos \dfrac{\theta }{2}}}$

Multiply numerator and denominator by $2\cos \dfrac{\theta }{2}$

$\Rightarrow \tan 20^\circ 30' = \dfrac{{2\sin \dfrac{\theta }{2}\cos \dfrac{\theta }{2}}}{{2\cos \dfrac{\theta }{2}\cos \dfrac{\theta }{2}}}$

$\Rightarrow \tan 20^\circ 30' = \dfrac{{2\sin \dfrac{\theta }{2}\cos \dfrac{\theta }{2}}}{{2{{\cos }^2}\dfrac{\theta }{2}}}$

We know that $\sin x = 2\sin \dfrac{x}{2}\cos \dfrac{x}{2}$ and $2{\cos ^2}\dfrac{x}{2} = 1 + \cos x$. Therefore, we get

$\Rightarrow \tan 20^\circ 30' = \dfrac{{\sin \theta }}{{1 + \cos \theta }}$

Put $\theta = 45^\circ$

$\Rightarrow \tan 20^\circ 30' = \dfrac{{\sin 45^\circ }}{{1 + \cos 45^\circ }}$

Substitute value of $\sin 45^\circ = \dfrac{1}{{\sqrt 2 }}$ and $\cos 45^\circ = \dfrac{1}{{\sqrt 2 }}$

$\Rightarrow \tan 20^\circ 30' = \dfrac{{\dfrac{1}{{\sqrt 2 }}}}{{1 + \dfrac{1}{{\sqrt 2 }}}}$

$\Rightarrow \tan 20^\circ 30' = \dfrac{{\dfrac{1}{{\sqrt 2 }}}}{{\dfrac{{\sqrt 2 + 1}}{{\sqrt 2 }}}}$

$\Rightarrow \tan 20^\circ 30' = \dfrac{1}{{\sqrt 2 + 1}}$

Multiply numerator and denominator by $\sqrt 2 - 1$

$\Rightarrow \tan 20^\circ 30' = \dfrac{1}{{\sqrt 2 + 1}} \times \dfrac{{\sqrt 2 - 1}}{{\sqrt 2 - 1}}$

$\Rightarrow \tan 20^\circ 30' = \dfrac{{\sqrt 2 - 1}}{{2 - 1}}$

$\Rightarrow \tan 20^\circ 30' = \sqrt 2 - 1$

This is our required answer.

1. Prove that $\sin 4A = 4\sin A{\cos ^3}A - 4\cos A{\sin ^3}A$.

Ans: We need to prove that $\sin 4A = 4\sin A{\cos ^3}A - 4\cos A{\sin ^3}A$

We can write $\sin 4A = \sin \left( {A + 3A} \right)$

As we know $\sin \left( {A + B} \right) = \sin A\cos B.\cos A\sin B$. Therefore, we get

$\Rightarrow \sin 4A = \sin A\cos 3A + \cos A\sin 3A$

As we know $\cos 3A = 4{\cos ^3}A - 3\cos A$ and $\sin 3A = 3\sin A - 4{\sin ^3}A$. Therefore, we get

$\Rightarrow \sin 4A = \sin A\left( {4{{\cos }^3}A - 3\cos A} \right) + \cos A\left( {3\sin A - 4{{\sin }^3}A} \right)$

On multiplication of terms, we get

$\Rightarrow \sin 4A = 4\sin A{\cos ^3}A - 3\sin A\cos A + 3\cos A\sin A - 4\cos A{\sin ^3}A$

$\Rightarrow \sin 4A = 4\sin A{\cos ^3}A - 4\cos A{\sin ^3}A$

Hence proved.

1.  If $\tan \theta + \sin \theta = m$ and $\tan \theta - \sin \theta = n$, then prove that ${m^2} - {n^2} = 4\sin \theta \tan \theta$.

Ans: Given, $\tan \theta + \sin \theta = m$ and $\tan \theta - \sin \theta = n$

$\Rightarrow \tan \theta + \sin \theta = m$

On squaring both the sides, we get

$\Rightarrow {\left( {\tan \theta + \sin \theta } \right)^2} = {m^2}$

As we know ${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$. Therefore, we get

$\Rightarrow {\tan ^2}\theta + {\sin ^2}\theta + 2\tan \theta \sin \theta = {m^2}....\left( i \right)$

$\Rightarrow \tan \theta - \sin \theta = n$

On squaring both the sides, we get

$\Rightarrow {\left( {\tan \theta - \sin \theta } \right)^2} = {n^2}$

As we know ${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$. Therefore, we get

$\Rightarrow {\tan ^2}\theta + {\sin ^2}\theta - 2\tan \theta \sin \theta = {n^2}....\left( {ii} \right)$

Now, we will subtract equation $\left( {ii} \right)$ from $\left( i \right)$.

$\Rightarrow {m^2} - {n^2} = {\tan ^2}\theta + {\sin ^2}\theta + 2\tan \theta \sin \theta - \left( {{{\tan }^2}\theta + {{\sin }^2}\theta - 2\tan \theta \sin \theta } \right)$

$\Rightarrow {m^2} - {n^2} = {\tan ^2}\theta + {\sin ^2}\theta + 2\tan \theta \sin \theta - {\tan ^2}\theta - {\sin ^2}\theta + 2\tan \theta \sin \theta$

On subtraction of terms, we get

$\Rightarrow {m^2} - {n^2} = 2\tan \theta \sin \theta + 2\tan \theta \sin \theta$

$\Rightarrow {m^2} - {n^2} = 4\sin \theta \tan \theta$

Hence proved.

1.  If $\tan \left( {A + B} \right) = p$, $\tan \left( {A - B} \right) = q$, then show that $\tan 2A = \dfrac{{p + q}}{{1 - pq}}$.

Ans: Given, $\tan \left( {A + B} \right) = p$ and $\tan \left( {A - B} \right) = q$

Let us start with LHS of $\tan 2A = \dfrac{{p + q}}{{1 - pq}}$.

$\Rightarrow \tan 2A = \tan \left( {A + B + A - B} \right)$

This can also be written as,

$\Rightarrow \tan 2A = \tan \left[ {\left( {A + B} \right) + \left( {A - B} \right)} \right]$

We know that $\tan \left( {x + y} \right) = \dfrac{{\tan x + \tan y}}{{1 - \tan x\tan y}}$. Now, we will apply this formula on the RHS.

$\Rightarrow \tan 2A = \dfrac{{\tan \left( {A + B} \right) + \tan \left( {A - B} \right)}}{{1 - \tan \left( {A + B} \right)\tan \left( {A - B} \right)}}$

We are given that $\tan \left( {A + B} \right) = p$ and $\tan \left( {A - B} \right) = q$. Therefore, we get

$\Rightarrow \tan 2A = \dfrac{{p + q}}{{1 - pq}}$

Hence proved.

1.  If $\cos \alpha + \cos \beta = 0 = \sin \alpha + \sin \beta$, then prove that $\cos 2\alpha + \cos 2\beta = - 2\cos \left( {\alpha + \beta } \right)$.

Ans: Given, $\cos \alpha + \cos \beta = 0$ and $\sin \alpha + \sin \beta = 0$.

$\Rightarrow {\left( {\cos \alpha + \cos \beta } \right)^2} = 0$

On squaring both the sides, we get

$\Rightarrow {\cos ^2}\alpha + {\cos ^2}\beta + 2\cos \alpha \cos \beta = 0........\left( 1 \right)$

$\Rightarrow \sin \alpha + \sin \beta = 0$

On squaring both the sides, we get

$\Rightarrow {\left( {\sin \alpha + \sin \beta } \right)^2} = 0$

$\Rightarrow {\sin ^2}\alpha + {\sin ^2}\beta + 2\sin \alpha \sin \beta = 0.......\left( 2 \right)$

Now, we will subtract equation $\left( 2 \right)$ from $\left( 1 \right)$.$\Rightarrow {\cos ^2}\alpha + {\cos ^2}\beta + 2\cos \alpha \cos \beta - \left( {{{\sin }^2}\alpha + {{\sin }^2}\beta + 2\sin \alpha \sin \beta } \right) = 0$

$\Rightarrow {\cos ^2}\alpha + {\cos ^2}\beta + 2\cos \alpha \cos \beta - {\sin ^2}\alpha - {\sin ^2}\beta - 2\sin \alpha \sin \beta = 0$

$\Rightarrow {\cos ^2}\alpha - {\sin ^2}\alpha + {\cos ^2}\beta - {\sin ^2}\beta + 2\cos \alpha \cos \beta - 2\sin \alpha \sin \beta = 0$

We know that ${\cos ^2}x - {\sin ^2}x = \cos 2x$. Therefore, we get

$\Rightarrow \cos 2\alpha + \cos 2\beta + 2\left( {\cos \alpha \cos \beta - \sin \alpha \sin \beta } \right) = 0$

We know that $\cos \left( {A + B} \right) = \cos A\cos B - \sin A\sin B$. Therefore, we get

$\Rightarrow \cos 2\alpha + \cos 2\beta + 2\cos \left( {\alpha + \beta } \right) = 0$

$\Rightarrow \cos 2\alpha + \cos 2\beta = - 2\cos \left( {\alpha + \beta } \right)$

Hence proved.

1.  If $\dfrac{{\sin \left( {x + y} \right)}}{{\sin \left( {x - y} \right)}} = \dfrac{{a + b}}{{a - b}}$, then show that $\dfrac{{\tan x}}{{\tan y}} = \dfrac{a}{b}$.

Ans: Given, $\dfrac{{\sin \left( {x + y} \right)}}{{\sin \left( {x - y} \right)}} = \dfrac{{a + b}}{{a - b}}$

We will apply componendo and dividend rule on the above written expression.

$\Rightarrow \dfrac{{\sin \left( {x + y} \right) + \sin \left( {x - y} \right)}}{{\sin \left( {x + y} \right) - \sin \left( {x - y} \right)}} = \dfrac{{a + b + a + b}}{{a + b - \left( {a - b} \right)}}$

We know that $\sin A + \sin B = 2\sin \dfrac{{A + B}}{2}.\cos \dfrac{{A - B}}{2}$ and $\sin A - \sin B = 2\cos \dfrac{{A + B}}{2}.\sin \dfrac{{A - B}}{2}$. Therefore, we get

$\Rightarrow \dfrac{{2\sin \left( {\dfrac{{x + y + x - y}}{2}} \right)\cos \left( {\dfrac{{x + y - x + y}}{2}} \right)}}{{2\cos \left( {\dfrac{{x + y + x - y}}{2}} \right)\sin \left( {\dfrac{{x + y - x + y}}{2}} \right)}} = \dfrac{{a + b + a + b}}{{a + b - a + b}}$

On simplification, we get

$\Rightarrow \dfrac{{2\sin \left( {\dfrac{{2x}}{2}} \right)\cos \left( {\dfrac{{2y}}{2}} \right)}}{{2\cos \left( {\dfrac{{2x}}{2}} \right)\sin \left( {\dfrac{{2y}}{2}} \right)}} = \dfrac{{2\left( {a + b} \right)}}{{2b}}$

On canceling common terms, we get

$\Rightarrow \dfrac{{\sin x\cos y}}{{\cos x\sin y}} = \dfrac{{a + b}}{b}$

Now we will write above written expression in terms of $\tan$ and $\cot$.

$\Rightarrow \tan x.\cot y = \dfrac{{a + b}}{b}$

$\Rightarrow \dfrac{{\tan x}}{{\tan y}} = \dfrac{{a + b}}{b}$

Hence proved.

1.  If $\tan \theta = \dfrac{{\sin \alpha - \cos \alpha }}{{\sin \alpha + \cos \alpha }}$, then show that $\sin \alpha + \cos \alpha = \sqrt 2 \cos \theta$.

Ans: Given, $\tan \theta = \dfrac{{\sin \alpha - \cos \alpha }}{{\sin \alpha + \cos \alpha }}$

Divide numerator and denominator of RHS by $\cos \alpha$.

$\Rightarrow \tan \theta = \dfrac{{\dfrac{{\sin \alpha - \cos \alpha }}{{\cos \alpha }}}}{{\dfrac{{\sin \alpha + \cos \alpha }}{{\cos \alpha }}}}$

$\Rightarrow \tan \theta = \dfrac{{\dfrac{{\sin \alpha }}{{\cos \alpha }} - \dfrac{{\cos \alpha }}{{\cos \alpha }}}}{{\dfrac{{\sin \alpha }}{{\cos \alpha }} + \dfrac{{\cos \alpha }}{{\cos \alpha }}}}$

As we know $\dfrac{{\sin x}}{{\cos x}} = \tan x$. Therefore, we get

$\Rightarrow \tan \theta = \dfrac{{\tan \alpha - 1}}{{\tan \alpha + 1}}$

We know that $\tan \dfrac{\pi }{4} = 1$. So, we can write above written equation as,

$\Rightarrow \tan \theta = \dfrac{{\tan \alpha - \tan \dfrac{\pi }{4}}}{{\tan \alpha + \tan \dfrac{\pi }{4}}}$

We know that $\tan \left( {x - y} \right) = \dfrac{{\tan x - \tan y}}{{1 + \tan x\tan y}}$. Therefore, we get

$\Rightarrow \tan \theta = \tan \left( {\alpha - \dfrac{\pi }{4}} \right)$

$\therefore \theta = \alpha - \dfrac{\pi }{4}$

$\Rightarrow \cos \theta = \cos \left( {\alpha - \dfrac{\pi }{4}} \right)$

We know that $\cos \left( {A - B} \right) = \cos A.\cos B + \sin A.\sin B$. Therefore, we get

$\Rightarrow \cos \theta = \cos \alpha .\cos \dfrac{\pi }{4} + \sin \alpha .\sin \dfrac{\pi }{4}$

Substitute values of $\cos \dfrac{\pi }{4} = \dfrac{1}{{\sqrt 2 }}$ and $\sin \dfrac{\pi }{4} = \dfrac{1}{{\sqrt 2 }}$

$\Rightarrow \cos \theta = \cos \alpha .\dfrac{1}{{\sqrt 2 }} + \sin \alpha .\dfrac{1}{{\sqrt 2 }}$

Multiply both sides by $\sqrt 2$.

$\Rightarrow \sqrt 2 \cos \theta = \cos \alpha + \sin \alpha$

$\Rightarrow \sin \alpha + \cos \alpha = \sqrt 2 \cos \theta$

Hence proved.

1.  If $\sin \theta + \cos \theta = 1$, then find the general value of $\theta$.

Ans: We have, $\sin \theta + \cos \theta = 1$

Divide both sides by $\sqrt 2$.

$\Rightarrow \dfrac{1}{{\sqrt 2 }}\sin \theta + \dfrac{1}{{\sqrt 2 }}\cos \theta = \dfrac{1}{{\sqrt 2 }}$

We know that $\dfrac{1}{{\sqrt 2 }} = \sin \dfrac{\pi }{4}$ and $\dfrac{1}{{\sqrt 2 }} = \cos \dfrac{\pi }{4}$. Therefore, we get

$\Rightarrow \sin \dfrac{\pi }{4}\sin \theta + \cos \dfrac{\pi }{4}\cos \theta = \dfrac{1}{{\sqrt 2 }}$

We know that $\cos \left( {A - B} \right) = \cos A\cos B + \sin A\sin B$. So, we can write above-written expression as

$\Rightarrow \cos \left( {\theta - \dfrac{\pi }{4}} \right) = \cos \dfrac{\pi }{4}$

We know that if $\cos \theta = \cos \alpha$, then $\theta = 2n\pi \pm \alpha$. Therefore, we get

$\Rightarrow \theta - \dfrac{\pi }{4} = 2n\pi \pm \dfrac{\pi }{4},n \in Z$

Transport $- \dfrac{\pi }{4}$ to RHS

$\Rightarrow \theta = 2n\pi \pm \dfrac{\pi }{4} - \dfrac{\pi }{4}$

$\Rightarrow \theta = 2n\pi + \dfrac{\pi }{4} + \dfrac{\pi }{4}$ or $\Rightarrow \theta = 2n\pi - \dfrac{\pi }{4} + \dfrac{\pi }{4}$

$\Rightarrow \theta = 2n\pi + \dfrac{\pi }{2}$ or $\Rightarrow \theta = 2n\pi ,n \in Z$

Therefore, the general values of $\theta$ are $2n\pi + \dfrac{\pi }{2}$ and $2n\pi$ where $n \in Z$.

1.  Find the most general value of $\theta$ satisfying the equation $\tan \theta = - 1$ and $\cos \theta = \dfrac{1}{{\sqrt 2 }}$.

Ans: We have, $\tan \theta = - 1$ and $\cos \theta = \dfrac{1}{{\sqrt 2 }}$.

As we can see $\tan \theta$ is negative and $\cos \theta$. It means $\theta$ lies in the fourth quadrant.

We have, $\tan \theta = - 1$

$\Rightarrow \tan \theta = \tan \left( { - \dfrac{\pi }{4}} \right)$

$\Rightarrow \tan \theta = \tan \left( { - \dfrac{\pi }{4}} \right)$

We know that $\tan \left( {2\pi - x} \right) = - \tan x$. So, we can write above written expression as

$\Rightarrow \tan \theta = \tan \left( {2\pi - \dfrac{\pi }{4}} \right)$

$\Rightarrow \tan \theta = \tan \left( {\dfrac{{7\pi }}{4}} \right)$

$\Rightarrow \theta = \dfrac{{7\pi }}{4}$

We have, $\cos \theta = \dfrac{1}{{\sqrt 2 }}$

$\Rightarrow \cos \theta = \cos \dfrac{\pi }{4}$

We know that $\cos \left( {2\pi - x} \right) = \cos x$. So, we can write above written expression as

$\Rightarrow \cos \theta = \cos \left( {2\pi - \dfrac{\pi }{4}} \right)$

$\Leftarrow \cos \theta = \cos \dfrac{{7\pi }}{4}$

$\Rightarrow \theta = \dfrac{{7\pi }}{4}$

Therefore, the general solution is $\theta = 2n\pi + \dfrac{{7\pi }}{4},n \in Z$.

1.  If $\cot \theta + \tan \theta = 2\cos ec\theta$, then find the general value of $\theta$.

Ans: Given, $\cot \theta + \tan \theta = 2\cos ec\theta$

Let us write above written expression in terms of $\sin$ and $\cos$.

$\Rightarrow \dfrac{{\cos \theta }}{{\sin \theta }} + \dfrac{{\sin \theta }}{{\cos \theta }} = \dfrac{2}{{\sin \theta }}$

Take LCM

$\Rightarrow \dfrac{{{{\cos }^2}\theta + {{\sin }^2}\theta }}{{\sin \theta \cos \theta }} = \dfrac{2}{{\sin \theta }}$

We know that ${\sin ^2}\theta + {\cos ^2}\theta = 1$. Therefore, we get

$\Rightarrow \dfrac{1}{{\sin \theta \cos \theta }} = \dfrac{2}{{\sin \theta }}$

On cross multiplication, we get

$\Rightarrow \dfrac{1}{{\sin \theta \cos \theta }} = \dfrac{2}{{\sin \theta }}$

$\Rightarrow 2\sin \theta \cos \theta = \sin \theta$

Transport $\sin \theta$ to LHS

$\Rightarrow 2\sin \theta \cos \theta - \sin \theta = 0$

$\Rightarrow \sin \theta \left( {2\cos \theta - 1} \right) = 0$

$\Rightarrow \sin \theta = 0$ or $2\cos \theta - 1 = 0$

$\Rightarrow \sin \theta = 0$ or $\cos \theta = \dfrac{1}{2}$

As we have, $\sin \theta = 0$

$\Rightarrow \theta = n\pi ,n \in Z$

As we have, $\cos \theta = \dfrac{1}{2}$

$\Rightarrow \cos \theta = \cos \dfrac{\pi }{3}$

$\Rightarrow \theta = 2n\pi \pm \dfrac{\pi }{3}$

Therefore, the general value of $\theta$ is $2n\pi \pm \dfrac{\pi }{3}$ and $n\pi$, $n \in Z$.

1.  If $2{\sin ^2}\theta = 3\cos \theta$, where $0 \leqslant \theta \leqslant 2\pi$, then find the value of $\theta$.

Ans: Given, $2{\sin ^2}\theta = 3\cos \theta$

We know that ${\sin ^2}x = 1 - {\cos ^2}x$. Therefore, we get

$\Rightarrow 2\left( {1 - {{\cos }^2}\theta } \right) = 3\cos \theta$

$\Rightarrow 2 - 2{\cos ^2}\theta = 3\cos \theta$

$\Rightarrow 2{\cos ^2}\theta + 3\cos \theta - 2 = 0$

$\Rightarrow 2{\cos ^2}\theta + 4\cos \theta - \cos \theta - 2 = 0$

$\Rightarrow 2\cos \theta \left( {\cos \theta + 2} \right) - 1\left( {\cos \theta + 2} \right) = 0$

$\Rightarrow \left( {\cos \theta + 2} \right)\left( {2\cos \theta - 1} \right) = 0$

$\Rightarrow \cos \theta + 2 = 0$ or $\Rightarrow 2\cos \theta - 1 = 0$

$\Rightarrow \cos \theta \ne - 2$ $\left[ { - 1 \leqslant \cos \theta \leqslant 1} \right]$

Therefore, $2\cos \theta - 1 = 0$

$\Rightarrow \cos \theta = \dfrac{1}{2}$

$\Rightarrow \cos \theta = \cos \dfrac{\pi }{3}$

$\Rightarrow \theta = \dfrac{\pi }{3}$ or $2\pi - \dfrac{\pi }{3}$

$\Rightarrow \theta = \dfrac{\pi }{3}$ or $\dfrac{{5\pi }}{3}$ ( As it is mentioned in the question that $0 \leqslant \theta \leqslant 2\pi$ )

Therefore, the value of $\theta$ are $\dfrac{\pi }{3}$ and $\dfrac{{5\pi }}{3}$.

1.  If $\sec x\cos 5x + 1 = 0$, where $0 < x \leqslant \dfrac{\pi }{2}$, then find the value of $x$.

Ans: Given, $\sec x\cos 5x + 1 = 0$

We can also write it as,

$\Rightarrow \dfrac{1}{{\cos x}}\cos 5x + 1 = 0$

Take LCM

$\Rightarrow \dfrac{{\cos 5x + \cos x}}{{\cos x}} = 0$

$\Rightarrow \cos 5x + \cos x = 0$

We know that $\cos C + \cos D = 2\cos \left( {\dfrac{{C + D}}{2}} \right)\cos \left( {\dfrac{{C - D}}{2}} \right)$. Therefore, we get

$\Rightarrow 2\cos \left( {\dfrac{{5x + x}}{2}} \right)\cos \left( {\dfrac{{5x - x}}{2}} \right) = 0$

$\Rightarrow \cos \left( {\dfrac{{6x}}{2}} \right)\cos \left( {\dfrac{{4x}}{2}} \right) = 0$

$\Rightarrow \cos 3x.\cos 2x = 0$

$\Rightarrow \cos 3x = 0$ or $\cos 2x = 0$

$\Rightarrow 3x = \dfrac{\pi }{2}$ or $2x = \dfrac{\pi }{2}$

$\Rightarrow x = \dfrac{\pi }{6}$ or $x = \dfrac{\pi }{4}$

Therefore, the value of $x$ are $\dfrac{\pi }{6}$, $\dfrac{\pi }{4}$.

Long Answer Type

1.  If $\sin \left( {\theta + \alpha } \right) = a$ and $\sin \left( {\theta + \beta } \right) = b$, then prove that $\cos 2\left( {\alpha - \beta } \right) - 4ab\cos \left( {\alpha - \beta } \right) = 1 - 2{a^2} - 2{b^2}$.

Ans: Given that, $\sin \left( {\theta + \alpha } \right) = a$ and $\sin \left( {\theta + \beta } \right) = b$.

We know that $\cos x = \sqrt {1 - {{\sin }^2}x}$. Therefore, we get

$\Rightarrow \cos \left( {\theta + \alpha } \right) = \sqrt {1 - {{\left( {\sin \left( {\theta + \alpha } \right)} \right)}^2}}$

$\Rightarrow \cos \left( {\theta + \alpha } \right) = \sqrt {1 - {a^2}}$

Similarly,

$\Rightarrow \cos \left( {\theta + \beta } \right) = \sqrt {1 - {{\left( {\sin \left( {\theta + \beta } \right)} \right)}^2}}$

$\Rightarrow \cos \left( {\theta + \beta } \right) = \sqrt {1 - {b^2}}$

Now, let us find value of $\cos \left( {\alpha - \beta } \right)$.

$\Rightarrow \cos \left( {\alpha - \beta } \right) = \cos \left[ {\left( {\theta + \alpha } \right) - \left( {\theta + \beta } \right)} \right]$

We know that $\cos \left( {A - B} \right) = \cos A.\cos B + \sin A.\sin B$. Therefore, we get

$\Rightarrow \cos \left( {\alpha - \beta } \right) = \cos \left( {\theta + \alpha } \right)\cos \left( {\theta + \beta } \right) + \sin \left( {\theta + \alpha } \right)\sin \left( {\theta + \beta } \right)$

On substituting the values, we get

$\Rightarrow \cos \left( {\alpha - \beta } \right) = \sqrt {1 - {a^2}} \sqrt {1 - {b^2}} + ab$

$\Rightarrow \cos \left( {\alpha - \beta } \right) = ab + \sqrt {1 - {b^2} - {a^2} + {a^2}{b^2}}$

Now, we will find the value of $\cos 2\left( {\alpha - \beta } \right)$

We know that $\cos 2A = 2{\cos ^2}A - 1$. Therefore, we get

$\Rightarrow \cos 2\left( {\alpha - \beta } \right) = 2{\cos ^2}\left( {\alpha - \beta } \right) - 1$

$\Rightarrow \cos 2\left( {\alpha - \beta } \right) = 2{\left( {ab + \sqrt {1 - {b^2} - {a^2} + {a^2}{b^2}} } \right)^2} - 1$

$\Rightarrow \cos 2\left( {\alpha - \beta } \right) = 2\left( {{a^2}{b^2} + 1 - {b^2} - {a^2} + {a^2}{b^2} + 2ab\sqrt {1 - {b^2} - {a^2} + {a^2}{b^2}} } \right) - 1$

We have, L.H.S. = $\cos 2\left( {\alpha - \beta } \right) - 4ab\cos \left( {\alpha - \beta } \right)$. On substituting the values, we get

$\Rightarrow 2\left( {{a^2}{b^2} + 1 - {b^2} - {a^2} + {a^2}{b^2} + 2ab\sqrt {1 - {b^2} - {a^2} + {a^2}{b^2}} } \right) - 1 - 4ab\left( {ab + \sqrt {1 - {b^2} - {a^2} + {a^2}{b^2}} } \right)$

$\Rightarrow 2{a^2}{b^2} + 2 - 2{b^2} - 2{a^2} + 2{a^2}{b^2} + 4ab\sqrt {1 - {b^2} - {a^2} + {a^2}{b^2}} - 1 - 4{a^2}{b^2} - 4ab\sqrt {1 - {b^2} - {a^2} + {a^2}{b^2}}$

$\Rightarrow 4{a^2}{b^2} + 1 - 2{a^2} - 2{b^2} + 4ab\sqrt {1 - {b^2} - {a^2} + {a^2}{b^2}} - 4{a^2}{b^2} - 4ab\sqrt {1 - {b^2} - {a^2} + {a^2}{b^2}}$

On subtraction, we get

$\Rightarrow 1 - 2{a^2} - 2{b^2}$

Hence proved.

1.  If $\cos \left( {\theta + \phi } \right) = m\cos \left( {\theta - \phi } \right)$, then prove that $\tan \theta = \dfrac{{1 - m}}{{1 + m}}\cot \phi$.

Ans: Given, $\cos \left( {\theta + \phi } \right) = m\cos \left( {\theta - \phi } \right)$

We can also write it as,

$\Rightarrow \dfrac{{\cos \left( {\theta + \phi } \right)}}{{\cos \left( {\theta - \phi } \right)}} = \dfrac{m}{1}$

Now, we will apply componendo and dividend rule on the above written expression.

$\Rightarrow \dfrac{{\cos \left( {\theta + \phi } \right) + \cos \left( {\theta - \phi } \right)}}{{\cos \left( {\theta + \phi } \right) - \cos \left( {\theta - \phi } \right)}} = \dfrac{{m + 1}}{{m - 1}}$

We know that $\cos A - \cos B = - 2\sin \dfrac{{A + B}}{2}.\sin \dfrac{{A - B}}{2}$ and $\cos A + \cos B = 2\cos \dfrac{{A + B}}{2}.\cos \dfrac{{A - B}}{2}$. Therefore, we get

$\Rightarrow \dfrac{{2\cos \left( {\dfrac{{\theta + \phi + \theta - \phi }}{2}} \right).\cos \left( {\dfrac{{\theta + \phi - \theta + \phi }}{2}} \right)}}{{ - 2\sin \left( {\dfrac{{\theta + \phi + \theta - \phi }}{2}} \right).\sin \left( {\dfrac{{\theta + \phi - \theta + \phi }}{2}} \right)}} = \dfrac{{m + 1}}{{m - 1}}$

$\Rightarrow \dfrac{{2\cos \left( {\dfrac{{2\theta }}{2}} \right).\cos \left( {\dfrac{{2\phi }}{2}} \right)}}{{ - 2\sin \left( {\dfrac{{2\theta }}{2}} \right).\sin \left( {\dfrac{{2\phi }}{2}} \right)}} = \dfrac{{m + 1}}{{m - 1}}$

On canceling common terms, we get

$\Rightarrow \dfrac{{\cos \left( \theta \right).\cos \left( \phi \right)}}{{ - \sin \left( \theta \right).\sin \left( \phi \right)}} = \dfrac{{m + 1}}{{m - 1}}$

As we know $\dfrac{{\cos x}}{{\sin x}} = \cot x$. So, we can write above-written equation as,

$\Rightarrow - \cot \theta .\cot \phi = \dfrac{{m + 1}}{{m - 1}}$

$\Rightarrow - \dfrac{{\cot \phi }}{{\tan \theta }} = \dfrac{{m + 1}}{{m - 1}}$

On cross multiplication, we get

$\Rightarrow \tan \theta \left( {1 + m} \right) = \cot \phi \left( {1 - m} \right)$

$\Rightarrow \tan \theta = \dfrac{{\left( {1 - m} \right)}}{{\left( {1 + m} \right)}}\cot \phi$

Hence proved.

1.  Find  the  value  of  the  expression  $3\left[ {{{\sin }^4}\left( {\dfrac{{3\pi }}{2} - \alpha } \right) + {{\sin }^4}\left( {3\pi + \alpha } \right)} \right] - 2\left[ {{{\sin }^6}\left( {\dfrac{\pi }{2} + \alpha } \right) + {{\sin }^6}\left( {5\pi - \alpha } \right)} \right]$

Ans: Given, $3\left[ {{{\sin }^4}\left( {\dfrac{{3\pi }}{2} - \alpha } \right) + {{\sin }^4}\left( {3\pi + \alpha } \right)} \right] - 2\left[ {{{\sin }^6}\left( {\dfrac{\pi }{2} + \alpha } \right) + {{\sin }^6}\left( {5\pi - \alpha } \right)} \right]$

We can also write it as,

$\Rightarrow 3\left[ {{{\sin }^4}\left( {\dfrac{{3\pi }}{2} - \alpha } \right) + {{\sin }^4}\left( {2\pi + \left( {\pi + \alpha } \right)} \right)} \right] - 2\left[ {{{\sin }^6}\left( {\dfrac{\pi }{2} + \alpha } \right) + {{\sin }^6}\left( {4\pi + \left( {\pi - \alpha } \right)} \right)} \right]$

As we know $\sin \left( {\dfrac{{3\pi }}{2} - x} \right) = - \cos x$, $\sin \left( {2\pi + x} \right) = \sin x$, $\sin \left( {\dfrac{\pi }{2} + x} \right) = \cos x$ and $\sin \left( {4\pi + x} \right) = \sin x$. Therefore, we get

$\Rightarrow 3\left[ {{{\cos }^4}\alpha + {{\sin }^4}\left( {\pi + \alpha } \right)} \right] - 2\left[ {{{\cos }^6}\alpha + {{\sin }^6}\left( {\pi - \alpha } \right)} \right]$

As we know $\sin \left( {\pi + \alpha } \right) = - \sin \alpha$ and $\sin \left( {\pi - \alpha } \right) = \sin \alpha$. Therefore, we get

$\Rightarrow 3\left[ {{{\cos }^4}\alpha + {{\sin }^4}\alpha } \right] - 2\left[ {{{\cos }^6}\alpha + {{\sin }^6}\alpha } \right]$

$\Rightarrow 3\left[ {{{\left( {{{\cos }^2}\alpha } \right)}^2} + {{\left( {{{\sin }^2}\alpha } \right)}^2}} \right] - 2\left[ {{{\left( {{{\cos }^2}\alpha } \right)}^3} + {{\left( {{{\sin }^2}\alpha } \right)}^3}} \right]..........\left( i \right)$

As we know ${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$. Therefore, we can write ${\left( {{{\cos }^2}\alpha } \right)^2} + {\left( {{{\sin }^2}\alpha } \right)^2}$ as ${\left( {{{\cos }^2}\alpha + {{\sin }^2}\alpha } \right)^2} - 2{\cos ^2}\alpha {\sin ^2}\alpha$. From here we can conclude that $\left( {{{\cos }^4}\alpha } \right) + \left( {{{\sin }^4}\alpha } \right) = {\left( {{{\cos }^2}\alpha + {{\sin }^2}\alpha } \right)^2} - 2{\cos ^2}\alpha {\sin ^2}\alpha ..........\left( {ii} \right)$

Also, as we know ${a^3} + {b^3} = \left( {a + b} \right)\left( {{a^2} + {b^2} - ab} \right)$. Therefore, we can write on expansion of  ${\left( {{{\cos }^2}\alpha } \right)^3} + {\left( {{{\sin }^2}\alpha } \right)^3}$, we get $\left( {{{\cos }^2}\alpha + {{\sin }^2}\alpha } \right)\left( {{{\cos }^4}\alpha + {{\sin }^4}\alpha - {{\cos }^2}\alpha {{\sin }^2}\alpha } \right)$.

Now, we will substitute these value in equation $\left( i \right)$.

$\Rightarrow 3\left[ {{{\left( {{{\cos }^2}\alpha + {{\sin }^2}\alpha } \right)}^2} - 2{{\cos }^2}\alpha {{\sin }^2}\alpha } \right] - 2\left[ {\left( {{{\cos }^2}\alpha + {{\sin }^2}\alpha } \right)\left( {{{\cos }^4}\alpha + {{\sin }^4}\alpha - {{\cos }^2}\alpha {{\sin }^2}\alpha } \right)} \right]$

As we know ${\sin ^2}x + {\cos ^2}x = 1$. Therefore, we get

$\Rightarrow 3\left( {1 - 2{{\cos }^2}\alpha {{\sin }^2}\alpha } \right) - 2\left( {{{\cos }^4}\alpha + {{\sin }^4}\alpha - {{\cos }^2}\alpha {{\sin }^2}\alpha } \right)$

From $\left( {ii} \right)$, we get

$\Rightarrow 3\left( {1 - 2{{\cos }^2}\alpha {{\sin }^2}\alpha } \right) - 2\left( {{{\left( {{{\cos }^2}\alpha + {{\sin }^2}\alpha } \right)}^2} - 2{{\cos }^2}\alpha {{\sin }^2}\alpha - {{\cos }^2}\alpha {{\sin }^2}\alpha } \right)$

As we know ${\sin ^2}x + {\cos ^2}x = 1$. Therefore, we get

$\Rightarrow 3\left( {1 - 2{{\cos }^2}\alpha {{\sin }^2}\alpha } \right) - 2\left( {1 - 2{{\cos }^2}\alpha {{\sin }^2}\alpha - {{\cos }^2}\alpha {{\sin }^2}\alpha } \right)$

$\Rightarrow 3\left( {1 - 2{{\cos }^2}\alpha {{\sin }^2}\alpha } \right) - 2\left( {1 - 3{{\cos }^2}\alpha {{\sin }^2}\alpha } \right)$

On multiplication, we get

$\Rightarrow 3 - 6{\cos ^2}\alpha {\sin ^2}\alpha - 2 + 6{\cos ^2}\alpha {\sin ^2}\alpha$

On subtraction, we get

$\Rightarrow 1$

Hence, value of $3\left[ {{{\sin }^4}\left( {\dfrac{{3\pi }}{2} - \alpha } \right) + {{\sin }^4}\left( {3\pi + \alpha } \right)} \right] - 2\left[ {{{\sin }^6}\left( {\dfrac{\pi }{2} + \alpha } \right) + {{\sin }^6}\left( {5\pi - \alpha } \right)} \right]$ is $1$.

1.  If $a\cos 2\theta + b\sin 2\theta = c$ has $\alpha$ and $\beta$ as its roots, then prove that $\tan \alpha + \tan \beta = \dfrac{{2b}}{{a + c}}$.

Ans: Given, $a\cos 2\theta + b\sin 2\theta = c$ has $\alpha$ and $\beta$ as its roots.

Let $a\cos 2\theta + b\sin 2\theta = c$ be equation $\left( i \right)$.

We know that $\cos 2x = \dfrac{{1 - {{\tan }^2}x}}{{1 + {{\tan }^2}x}}$ and  $\sin 2x = \dfrac{{2\tan x}}{{1 + {{\tan }^2}x}}$. Therefore, we get

$\Rightarrow a\left( {\dfrac{{1 - {{\tan }^2}\theta }}{{1 + {{\tan }^2}\theta }}} \right) + b\left( {\dfrac{{2\tan \theta }}{{1 + {{\tan }^2}\theta }}} \right) = c$

Take LCM

$\Rightarrow \dfrac{{a\left( {1 - {{\tan }^2}\theta } \right) + b\left( {2\tan \theta } \right)}}{{1 + {{\tan }^2}\theta }} = c$

On cross multiplication, we get

$\Rightarrow a\left( {1 - {{\tan }^2}\theta } \right) + b\left( {2\tan \theta } \right) = c\left( {1 + {{\tan }^2}\theta } \right)$

On multiplication of terms, we get

$\Rightarrow a - a{\tan ^2}\theta + 2b\tan \theta = c + c{\tan ^2}\theta$

$\Rightarrow a - a{\tan ^2}\theta + 2b\tan \theta - c - c{\tan ^2}\theta = 0$

Divide whole equation by $- 1$.

$\Rightarrow - a + a{\tan ^2}\theta - 2b\tan \theta + c + c{\tan ^2}\theta = 0$

$\Rightarrow \left( {a + c} \right){\tan ^2}\theta - 2b\tan \theta + \left( {c - a} \right) = 0.......\left( {ii} \right)$

We are given that $\alpha$ and $\beta$ are the roots of equation $\left( i \right)$. Thus,  $\tan \alpha$ and $\tan \beta$ are the roots of the equation $\left( {ii} \right)$.

As we know sum of roots of a quadratic equation $a{x^2} + bx + c = 0$ is $\dfrac{{ - b}}{a}$. Therefore, we get

$\Rightarrow \tan \alpha + \tan \beta = \dfrac{{ - \left( { - 2b} \right)}}{{a + c}}$

$\Rightarrow \tan \alpha + \tan \beta = \dfrac{{2b}}{{a + c}}$

Hence proved.

1.  If $x = \sec \phi - \tan \phi$ and $y = \cos ec\phi + \cot \phi$ then show that $xy + x - y + 1 = 0$.

Ans: Given, $x = \sec \phi - \tan \phi$ and $y = \cos ec\phi + \cot \phi$

We have, L.H.S. = $xy + x - y + 1$.

Now, we will substitute the given values in above written expression.

$\Rightarrow \left( {\sec \phi - \tan \phi } \right)\left( {\cos ec\phi + \cot \phi } \right) + \left( {\sec \phi - \tan \phi } \right) - \left( {\cos ec\phi + \cot \phi } \right) + 1$

Now, we will write above written expression in terms of $\sin$ and $\cos$.

$\Rightarrow \left( {\dfrac{{1 - \sin \phi }}{{\cos \phi }}} \right)\left( {\dfrac{{1 + \cos \phi }}{{\sin \phi }}} \right) + \left( {\dfrac{{1 - \sin \phi }}{{\cos \phi }}} \right) - \left( {\dfrac{{1 + \cos \phi }}{{\sin \phi }}} \right) + 1$

$\Rightarrow \dfrac{{\left( {1 - \sin \phi } \right)\left( {1 + \cos \phi } \right)}}{{\cos \phi \sin \phi }} + \left( {\dfrac{{\sin \phi \left( {1 - \sin \phi } \right) - \cos \phi \left( {1 + \cos \phi } \right)}}{{\cos \phi \sin \phi }}} \right) + 1$

$\Rightarrow \dfrac{{1 - \sin \phi + \cos \phi - \sin \phi \cos \phi }}{{\cos \phi \sin \phi }} + \left( {\dfrac{{\sin \phi - {{\sin }^2}\phi - \cos \phi - {{\cos }^2}\phi }}{{\cos \phi \sin \phi }}} \right) + 1$

$\Rightarrow \dfrac{{1 - \sin \phi + \cos \phi - \sin \phi \cos \phi + \sin \phi - \cos \phi - \left( {{{\sin }^2}\phi + {{\cos }^2}\phi } \right) + \sin \phi \cos \phi }}{{\cos \phi \sin \phi }}$

We know that ${\sin ^2}x + {\cos ^2}x = 1$. Therefore, we get

$\Rightarrow \dfrac{{1 - 1}}{{\cos \phi \sin \phi }} = 0$

Hence proved.

1.  If $\theta$ lies in the first quadrant and $\cos \theta = \dfrac{8}{{17}}$, then find the value of $\cos \left( {30^\circ + \theta } \right) + \cos \left( {45^\circ - \theta } \right) + \cos \left( {120^\circ - \theta } \right)$.

Ans: Here, we are given that $\cos \theta = \dfrac{8}{{17}}$, $\theta$ lies in the first quadrant.

We know that $\sin x = \sqrt {1 - {{\cos }^2}x}$. Therefore, we get

$\Rightarrow \sin \theta = \sqrt {1 - {{\left( {\dfrac{8}{{17}}} \right)}^2}}$

$\Rightarrow \sin \theta = \sqrt {1 - \dfrac{{64}}{{289}}}$

$\Rightarrow \sin \theta = \sqrt {\dfrac{{289 - 64}}{{289}}}$

$\Rightarrow \sin \theta = \sqrt {\dfrac{{225}}{{289}}}$

$\Rightarrow \sin \theta = \pm \dfrac{{15}}{{17}}$

But $\theta$ lies in the first quadrant and in first quadrant $\sin e$ is positive.

$\Rightarrow \sin \theta = \dfrac{{15}}{{17}}$

We need to find the value of $\cos \left( {30^\circ + \theta } \right) + \cos \left( {45^\circ - \theta } \right) + \cos \left( {120^\circ - \theta } \right)$.

We know that $\cos \left( {A + B} \right) = \cos A\cos B - \sin A\sin B$ and $\cos \left( {A - B} \right) = \cos A\cos B + \sin A\sin B$. Therefore, we get

$\Rightarrow \cos 30^\circ \cos \theta - \sin 30^\circ \sin \theta + \cos 45^\circ \cos \theta + \sin 45^\circ \sin \theta + \cos 120^\circ \cos \theta + \sin 120^\circ \sin \theta$

Substitute value of $\cos 30^\circ = \dfrac{{\sqrt 3 }}{2}$, $\sin 30^\circ = \dfrac{1}{2}$, $\cos 45^\circ = \dfrac{1}{{\sqrt 2 }}$, $\sin 45^\circ = \dfrac{1}{{\sqrt 2 }}$, $\cos 120^\circ = - \dfrac{1}{2}$ and $\sin 120^\circ = \dfrac{{\sqrt 3 }}{2}$.

$\Rightarrow \dfrac{{\sqrt 3 }}{2} \times \cos \theta - \dfrac{1}{2} \times \sin \theta + \dfrac{1}{{\sqrt 2 }} \times \cos \theta + \dfrac{1}{{\sqrt 2 }} \times \sin \theta - \dfrac{1}{2}\cos \theta + \dfrac{{\sqrt 3 }}{2}\sin \theta$

$\Rightarrow \dfrac{{\sqrt 3 }}{2}\left( {\cos \theta + \sin \theta } \right) - \dfrac{1}{2}\left( {\cos \theta + \sin \theta } \right) + \dfrac{1}{{\sqrt 2 }}\left( {\cos \theta + \sin \theta } \right)$

$\Rightarrow \left( {\dfrac{{\sqrt 3 }}{2} - \dfrac{1}{2} + \dfrac{1}{{\sqrt 2 }}} \right)\left( {\cos \theta + \sin \theta } \right)$

$\Rightarrow \left( {\dfrac{{\sqrt 3 - 1}}{2} + \dfrac{1}{{\sqrt 2 }}} \right)\left( {\dfrac{8}{{17}} + \dfrac{{15}}{{17}}} \right)$

$\Rightarrow \left( {\dfrac{{\sqrt 3 - 1 + \sqrt 2 }}{2}} \right)\left( {\dfrac{{23}}{{17}}} \right)$

$\Rightarrow \dfrac{{23}}{{34}}\left( {\sqrt 3 - 1 + \sqrt 2 } \right)$

This is our required answer.

1.  Find the value of the expression ${\cos ^4}\dfrac{\pi }{8} + {\cos ^4}\dfrac{{3\pi }}{8} + {\cos ^4}\dfrac{{5\pi }}{8} + {\cos ^4}\dfrac{{7\pi }}{8}$.

Ans: Given expression, ${\cos ^4}\dfrac{\pi }{8} + {\cos ^4}\dfrac{{3\pi }}{8} + {\cos ^4}\dfrac{{5\pi }}{8} + {\cos ^4}\dfrac{{7\pi }}{8}$

This can also be written as,

$\Rightarrow {\cos ^4}\dfrac{\pi }{8} + {\cos ^4}\dfrac{{3\pi }}{8} + {\cos ^4}\left( {\pi - \dfrac{{3\pi }}{8}} \right) + {\cos ^4}\left( {\pi - \dfrac{\pi }{8}} \right)$

We know that $\cos \left( {\pi - \theta } \right) = - \cos \theta$. Therefore, we get

$\Rightarrow {\cos ^4}\dfrac{\pi }{8} + {\cos ^4}\dfrac{{3\pi }}{8} + {\cos ^4}\dfrac{{3\pi }}{8} + {\cos ^4}\dfrac{\pi }{8}$

(Here, ${\cos ^4}\dfrac{{3\pi }}{8}$, ${\cos ^4}\dfrac{\pi }{8}$ are positive because power is even)

$\Rightarrow 2{\cos ^4}\dfrac{\pi }{8} + 2{\cos ^4}\dfrac{{3\pi }}{8}$

$\Rightarrow 2\left[ {{{\cos }^4}\dfrac{\pi }{8} + {{\cos }^4}\dfrac{{3\pi }}{8}} \right]$

$\Rightarrow 2\left[ {{{\cos }^4}\dfrac{\pi }{8} + {{\cos }^4}\left( {\dfrac{\pi }{2} - \dfrac{\pi }{8}} \right)} \right]$

We know that $\cos \left( {\dfrac{\pi }{2} - \theta } \right) = \sin \theta$. Therefore, we get

$\Rightarrow 2\left[ {{{\cos }^4}\dfrac{\pi }{8} + {{\sin }^4}\dfrac{\pi }{8}} \right]$

We know that ${a^2} + {b^2} = {\left( {a + b} \right)^2} - 2ab$. Therefore, we get

$\Rightarrow 2\left[ {{{\left( {{{\cos }^2}\dfrac{\pi }{8} + {{\sin }^2}\dfrac{\pi }{8}} \right)}^2} - 2{{\sin }^2}\dfrac{\pi }{8} \times {{\cos }^2}\dfrac{\pi }{8}} \right]$

$\Rightarrow 2\left[ {1 - 2{{\sin }^2}\dfrac{\pi }{8} \times {{\cos }^2}\dfrac{\pi }{8}} \right]$

$\Rightarrow 2 - 4{\sin ^2}\dfrac{\pi }{8} \times {\cos ^2}\dfrac{\pi }{8}$

$\Rightarrow 2 - {\left( {2\sin \dfrac{\pi }{8} \times \cos \dfrac{\pi }{8}} \right)^2}$

We know that $\sin 2x = 2\sin x\cos x$. Therefore, we get

$\Rightarrow 2 - {\left( {\sin \dfrac{\pi }{4}} \right)^2}$

Substitute $\sin \dfrac{\pi }{4} = \dfrac{1}{{\sqrt 2 }}$.

$\Rightarrow 2 - {\left( {\dfrac{1}{{\sqrt 2 }}} \right)^2}$

$\Rightarrow 2 - \dfrac{1}{2} = \dfrac{{4 - 1}}{2}$

$\Rightarrow \dfrac{3}{2}$

Therefore, value of given expression is $\dfrac{3}{2}$.

1.  Find the general solution of the equation $5{\cos ^2}\theta + 7{\sin ^2}\theta - 6 = 0$.

Ans: Given, $5{\cos ^2}\theta + 7{\sin ^2}\theta - 6 = 0$

We know that ${\sin ^2}x = 1 - {\cos ^2}x$. Therefore, we get

$\Rightarrow 5{\cos ^2}\theta + 7\left( {1 - {{\cos }^2}x} \right) - 6 = 0$

$\Rightarrow 5{\cos ^2}\theta + 7 - 7{\cos ^2}x - 6 = 0$

$\Rightarrow 1 - 2{\cos ^2}x = 0$

$\Rightarrow 2{\cos ^2}x = 1$

$\Rightarrow {\cos ^2}x = \dfrac{1}{2}$

$\Rightarrow {\cos ^2}x = \cos \dfrac{\pi }{4}$

We know that $1 + \cos 2x = 2{\cos ^2}x$. Therefore, we get

$\Rightarrow \dfrac{{1 + \cos 2\theta }}{2} = \dfrac{{1 + \cos 2 \times \dfrac{\pi }{4}}}{2}$

$\Rightarrow 1 + \cos 2\theta = 1 + \cos \dfrac{\pi }{2}$

$\Rightarrow \cos 2\theta = \cos \dfrac{\pi }{2}$

We know that if $\cos \theta = \cos \alpha$ then $\theta = 2n\pi \pm \alpha$, $n \in Z$

$\Rightarrow 2\theta = 2n\pi \pm \dfrac{\pi }{2}$

$\Rightarrow \theta = n\pi \pm \dfrac{\pi }{4}$

Therefore, the general solution given equation is $\theta = n\pi \pm \dfrac{\pi }{4}$ where $n \in Z$.

1.  Find the general solution of the equation $\sin x - 3\sin 2x + \sin 3x = \cos x - 3\cos 2x + \cos 3x$.

Ans: Given equation, $\sin x - 3\sin 2x + \sin 3x = \cos x - 3\cos 2x + \cos 3x$

$\Rightarrow \left( {\sin 3x + \sin x} \right) - 3\sin 2x = \left( {\cos 3x + \cos x} \right) - 3\cos 2x$

We know that $\sin A + \sin B = 2\sin \dfrac{{A + B}}{2}\cos \dfrac{{A - B}}{2}$ and $\cos A + \cos B = 2\cos \dfrac{{A + B}}{2}\cos \dfrac{{A - B}}{2}$. Therefore, we get

$\Rightarrow 2\sin \dfrac{{3x + x}}{2}\cos \dfrac{{3x - x}}{2} - 3\sin 2x = 2\cos \dfrac{{3x + x}}{2}\cos \dfrac{{3x - x}}{2} - 3\cos 2x$

$\Rightarrow 2\sin 2x.\cos x - 3\sin 2x = 2\cos 2x.\cos x - 3\cos 2x$

$\Rightarrow 2\cos x\left( {\sin 2x - \cos 2x} \right) = 3\left( {\sin 2x - \cos 2x} \right)$

$\Rightarrow 2\cos x\left( {\sin 2x - \cos 2x} \right) - 3\left( {\sin 2x - \cos 2x} \right) = 0$

$\Rightarrow \left( {\sin 2x - \cos 2x} \right)\left( {2\cos x - 3} \right) = 0$

$\Rightarrow \sin 2x - \cos 2x = 0$ or $\Rightarrow 2\cos x - 3 \ne 0$ ( $- 1 \leqslant \cos x \leqslant 1$ )

$\Rightarrow \sin 2x = \cos 2x$

$\Rightarrow \dfrac{{\sin 2x}}{{\cos 2x}} = 1$

$\Rightarrow \tan 2x = 1$

We know that if $\tan \theta = \tan \alpha$ then $\theta = n\pi + \alpha$, $n \in Z$

$\Rightarrow \tan 2x = \tan \dfrac{\pi }{4}$

$\Rightarrow 2x = n\pi + \dfrac{\pi }{4}$

$\Rightarrow x = \dfrac{{n\pi }}{2} + \dfrac{\pi }{8}$

Therefore, the general solution given equation is $x = \dfrac{{n\pi }}{2} + \dfrac{\pi }{8}$ where $n \in Z$.

1.  Find the general solution of the equation $\left( {\sqrt 3 - 1} \right)\cos \theta + \left( {\sqrt 3 + 1} \right)\sin \theta = 2$.

Ans: Given, $\left( {\sqrt 3 - 1} \right)\cos \theta + \left( {\sqrt 3 + 1} \right)\sin \theta = 2$

Put $\sqrt 3 - 1 = r\sin \alpha ......\left( i \right)$ and $\sqrt 3 + 1 = r\cos \alpha ......\left( {ii} \right)$

On squaring and adding both the sides, we get

$\Rightarrow {\left( {r\sin \alpha } \right)^2} + {\left( {r\cos \alpha } \right)^2} = {\left( {\sqrt 3 - 1} \right)^2} + {\left( {\sqrt 3 + 1} \right)^2}$

$\Rightarrow {r^2}{\sin ^2}\alpha + {r^2}{\cos ^2}\alpha = 3 + 1 - 2\sqrt 3 + 3 + 1 + 2\sqrt 3$

$\Rightarrow {r^2}\left( {{{\sin }^2}\alpha + {{\cos }^2}\alpha } \right) = 8$

We know that ${\sin ^2}x + {\cos ^2}x = 1$. Therefore, we get

$\Rightarrow {r^2} = 8$

$\Rightarrow r = 2\sqrt 2$

The given equation can be written as,

$\Rightarrow r\sin \alpha .\cos \theta + r\cos \alpha .\sin \theta = 2$

$\Rightarrow r\left( {\sin \alpha .\cos \theta + \cos \alpha .\sin \theta } \right) = 2$

We know that $\sin \left( {A + B} \right) = \sin A\cos B + \cos A\sin B$. Therefore, we get

$\Rightarrow 2\sqrt 2 \sin \left( {\alpha + \theta } \right) = 2$

$\Rightarrow \sqrt 2 \sin \left( {\alpha + \theta } \right) = 1$

$\Rightarrow \sin \left( {\alpha + \theta } \right) = \dfrac{1}{{\sqrt 2 }}$

$\Rightarrow \sin \left( {\alpha + \theta } \right) = \sin \dfrac{\pi }{4}$

We know that if $\sin \theta = \sin \alpha$ then $\theta = n\pi + {\left( { - 1} \right)^n}\alpha$, $n \in Z$

$\Rightarrow \alpha + \theta = n\pi + {\left( { - 1} \right)^n}.\dfrac{\pi }{4}.......\left( {iii} \right)$

We have, $\dfrac{{r\sin \alpha }}{{r\cos \alpha }} = \dfrac{{\sqrt 3 - 1}}{{\sqrt 3 + 1}}$ from equation $\left( i \right)$ and $\left( {ii} \right)$.

This can also be written as, $\tan \alpha = \dfrac{{\sqrt 3 - 1}}{{1 + \sqrt 3 .1}}$

We know that $\tan \dfrac{\pi }{3} = \sqrt 3$ and $\tan \dfrac{\pi }{4} = 1$. Therefore, we get

$\Rightarrow \tan \alpha = \dfrac{{\tan \dfrac{\pi }{3} - \tan \dfrac{\pi }{4}}}{{1 + \tan \dfrac{\pi }{3}.\tan \dfrac{\pi }{4}}}$

We know that $\tan \left( {x - y} \right) = \dfrac{{\tan x - \tan y}}{{1 + \tan x.\tan y}}$. Therefore, we get

$\Rightarrow \tan \alpha = \tan \left( {\dfrac{\pi }{3} - \dfrac{\pi }{4}} \right)$

$\Rightarrow \tan \alpha = \tan \dfrac{\pi }{{12}}$

$\Rightarrow \alpha = \dfrac{\pi }{{12}}$

Now, we will substitute value of $\alpha$, in equation $\left( {iii} \right)$.

$\Rightarrow \dfrac{\pi }{{12}} + \theta = n\pi + {\left( { - 1} \right)^n}.\dfrac{\pi }{4}$

$\Rightarrow \theta = n\pi + {\left( { - 1} \right)^n}.\dfrac{\pi }{4} - \dfrac{\pi }{{12}}$

Therefore, the general solution given equation is $\theta = n\pi + {\left( { - 1} \right)^n}.\dfrac{\pi }{4} - \dfrac{\pi }{{12}}$ where $n \in Z$.

OBJECTIVE TYPE QUESTIONS

Choose the correct answer from the given options in the exercises $30$ to $59$.

1.  If $\sin \theta + \cos ec\theta = 2$, then ${\sin ^2}\theta + \cos e{c^2}\theta$ is equal to

1. $1$

2. $4$

3. $2$

4. None of these

Ans: Given, $\sin \theta + \cos ec\theta = 2$

On squaring both the sides, we get

$\Rightarrow {\left( {\sin \theta + \cos ec\theta } \right)^2} = {2^2}$

We know that ${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$. Therefore, we get

$\Rightarrow {\sin ^2}\theta + \cos e{c^2}\theta + 2\sin \theta \cos ec\theta = 4$

$\Rightarrow {\sin ^2}\theta + \cos e{c^2}\theta + 2\sin \theta \dfrac{1}{{\sin \theta }} = 4$

On canceling common terms, we get

$\Rightarrow {\sin ^2}\theta + \cos e{c^2}\theta + 2 = 4$

Transport $2$ to the RHS

$\Rightarrow {\sin ^2}\theta + \cos e{c^2}\theta = 4 - 2$

$\Rightarrow {\sin ^2}\theta + \cos e{c^2}\theta = 2$

Hence, option (c) is the correct answer.

1.  If $f\left( x \right) = {\cos ^2}x + {\sec ^2}x$, then

1. $f\left( x \right) < 1$

2. $f\left( x \right) = 1$

3. $2 < f\left( x \right) < 1$

4. $f\left( x \right) \geqslant 2$

Ans: Given, $f\left( x \right) = {\cos ^2}x + {\sec ^2}x$

As we know $AM \geqslant GM$. Therefore, we get

$\Rightarrow \dfrac{{{{\cos }^2}x + {{\sec }^2}x}}{2} \geqslant \sqrt {{{\cos }^2}x \times {{\sec }^2}x}$

We know that $\cos x = \dfrac{1}{{\sec x}}$. Therefore, we get

$\Rightarrow \dfrac{{{{\cos }^2}x + {{\sec }^2}x}}{2} \geqslant 1$

On cross multiplication, we get

$\Rightarrow {\cos ^2}x + {\sec ^2}x \geqslant 2$

$\Rightarrow f\left( x \right) \geqslant 2$

Hence, option (d) is the correct answer.

1.  If $\tan \theta = \dfrac{1}{2}$ and $\tan \phi = \dfrac{1}{3}$, then the value of $\theta + \phi$ is

1. $\dfrac{\pi }{6}$

2. $\pi$

3. $0$

4. $\dfrac{\pi }{4}$

Ans: Given, $\tan \theta = \dfrac{1}{2}$ and $\tan \phi = \dfrac{1}{3}$

We know that $\tan \left( {\theta + \phi } \right) = \dfrac{{\tan \theta + \tan \phi }}{{1 - \tan \theta \tan \phi }}$. Therefore, we will substitute the given values in this formula.

$\Rightarrow \tan \left( {\theta + \phi } \right) = \dfrac{{\dfrac{1}{2} + \dfrac{1}{3}}}{{1 - \dfrac{1}{2} \times \dfrac{1}{3}}}$

On taking LCM, we get

$\Rightarrow \tan \left( {\theta + \phi } \right) = \dfrac{{\dfrac{{3 + 2}}{6}}}{{\dfrac{{6 - 1}}{6}}}$

$\Rightarrow \tan \left( {\theta + \phi } \right) = \dfrac{{\dfrac{5}{6}}}{{\dfrac{5}{6}}} = 1$

As we know $\tan \dfrac{\pi }{4} = 1$. Therefore, we get

$\Rightarrow \tan \left( {\theta + \phi } \right) = \tan \dfrac{\pi }{4}$

$\Rightarrow \theta + \phi = \dfrac{\pi }{4}$

Hence, option (d) is the correct answer.

1.  Which of the following is not correct?

1. $\sin \theta = - \dfrac{1}{5}$

2. $\cos \theta = 1$

3. $\sec \theta = \dfrac{1}{2}$

4. $\tan \theta = 20$

Ans:

As we know $- 1 \leqslant \sin \theta \leqslant 1$. Therefore, $\sin \theta = - \dfrac{1}{5}$ is correct.

We know that $\cos 0^\circ = 1$. Therefore, $\cos \theta = 1$ is correct.

We have, $\sec \theta = \dfrac{1}{2}$. This can also be written as

$\Rightarrow \cos \theta = 2$

Which is not correct because $- 1 \leqslant \sin \theta \leqslant 1$.

Hence, option (d) is the correct answer.

1. The value of $\tan 1^\circ \tan 2^\circ \tan 3^\circ .....\tan 89^\circ$ is

1. $0$

2. $1$

3. $\dfrac{1}{2}$

4. Not defined

Ans: We need to find the value of $\tan 1^\circ \tan 2^\circ \tan 3^\circ .....\tan 89^\circ$

$\Rightarrow \tan 1^\circ \tan 2^\circ .....\tan 44^\circ \tan 45^\circ \tan \left( {90^\circ - 44^\circ } \right).......\tan \left( {90^\circ - 2^\circ } \right)\tan \left( {90^\circ - 1^\circ } \right)$

We know that $\tan \left( {90^\circ - \theta } \right) = \cot \theta$. Therefore, we get

$\Rightarrow \left[ {\tan 1^\circ \tan 2^\circ .....\tan 44^\circ } \right]\tan 45^\circ \left[ {\cot 44^\circ ......\cot 2^\circ .\cot 1^\circ } \right]$

$\Rightarrow \left[ {\left( {\tan 1^\circ \times \cot 1^\circ } \right)\left( {\tan 2^\circ \cot 2^\circ } \right).....\left( {\tan 44^\circ \cot 44^\circ } \right)} \right] \times \tan 45^\circ$

We know that $\tan A \times \cot A = 1$. Therefore, we get

$\Rightarrow 1 \times 1... \times 1 \times \tan 45^\circ$

Substitute value of $\tan 45^\circ = 1$.

$\Rightarrow 1 \times 1... \times 1 \times 1 = 1$

Hence, option (b) is the correct answer.

1.  The value of $\dfrac{{1 - {{\tan }^2}15^\circ }}{{1 + {{\tan }^2}15^\circ }}$ is

1. $1$

2. $\sqrt 3$

3. $\dfrac{{\sqrt 3 }}{2}$

4. $2$

Ans: We have to find the value of $\dfrac{{1 - {{\tan }^2}15^\circ }}{{1 + {{\tan }^2}15^\circ }}$

Let $\theta = 15^\circ$

$\Rightarrow 2\theta = 30^\circ$

As we know $\cos 2\theta = \dfrac{{1 - {{\tan }^2}\theta }}{{1 + {{\tan }^2}\theta }}$. Therefore, on substituting value of $\theta$ we get

$\Rightarrow \cos 30^\circ = \dfrac{{1 - {{\tan }^2}15^\circ }}{{1 + {{\tan }^2}15^\circ }}$

Substitute value of $\tan 30^\circ = \dfrac{{\sqrt 3 }}{2}$.

$\Rightarrow \dfrac{{\sqrt 3 }}{2} = \dfrac{{1 - {{\tan }^2}15^\circ }}{{1 + {{\tan }^2}15^\circ }}$

Or

$\Rightarrow \dfrac{{1 - {{\tan }^2}15^\circ }}{{1 + {{\tan }^2}15^\circ }} = \dfrac{{\sqrt 3 }}{2}$

Hence, option (c) is the correct answer.

1.  The value of $\cos 1^\circ \cos 2^\circ \cos 3^\circ .....\cos 179^\circ$ is

1. $\dfrac{1}{{\sqrt 2 }}$

2. $0$

3. $1$

4. $- 1$

Ans: Given expression: $\cos 1^\circ \cos 2^\circ \cos 3^\circ .....\cos 179^\circ$

$\Rightarrow \cos 1^\circ \cos 2^\circ \cos 3^\circ .....\cos 90^\circ ......\cos 179^\circ$

As we know $\cos 90^\circ = 0$. Therefore, we get

$\Rightarrow \cos 1^\circ \cos 2^\circ \cos 3^\circ ..... \times 0 \times ......\cos 179^\circ$

$\Rightarrow 0$

Hence, option (b) is the correct answer.

1.  If $\tan \theta = 3$ and $\theta$ lies in third quadrant, then the value of $\sin \theta$ is

1. $\dfrac{1}{{\sqrt {10} }}$

2. $- \dfrac{1}{{\sqrt {10} }}$

3. $\dfrac{{ - 3}}{{\sqrt {10} }}$

4. $\dfrac{3}{{\sqrt {10} }}$

Ans: Given, $\tan \theta = 3$

$\Rightarrow \tan \theta = \dfrac{P}{B} = \dfrac{3}{1}$

$\Rightarrow H = \sqrt {{3^2} + {1^2}}$

$\Rightarrow H = \sqrt {9 + 1}$

$\Rightarrow H = \sqrt {10}$

$\Rightarrow \sin \theta = - \dfrac{3}{{\sqrt {10} }}$

( The value of $\sin \theta$ is negative because $\theta$ lies in the third quadrant )

Hence, option (a) is the correct answer.

1.  The value of $\tan 75^\circ - \cot 75^\circ$ is equal to

1. $2\sqrt 3$

2. $2 + \sqrt 3$

3. $2 - \sqrt 3$

4. $1$

Ans: Given expression, $\tan 75^\circ - \cot 75^\circ$

Above written expression can also be written as,

$\Rightarrow \tan 75^\circ - \cot \left( {90^\circ - 15^\circ } \right)$

We know that $\cot \left( {90^\circ - \theta } \right) = \tan \theta$. Therefore, we get

$\Rightarrow \tan 75^\circ - \tan 15^\circ$

$\Rightarrow \dfrac{{\sin 75^\circ }}{{\cos 75^\circ }} - \dfrac{{\sin 15^\circ }}{{\cos 15^\circ }}$

Take LCM

$\Rightarrow \dfrac{{\sin 75^\circ \cos 15^\circ - \cos 75^\circ \sin 15^\circ }}{{\cos 75^\circ \cos 15^\circ }}$

We know that $\sin \left( {A - B} \right) = \sin A\cos B - \cos A\sin B$. Therefore, we get

$\Rightarrow \dfrac{{\sin \left( {75^\circ - 15^\circ } \right)}}{{\cos 75^\circ \cos 15^\circ }}$

$\Rightarrow \dfrac{{\sin \left( {75^\circ - 15^\circ } \right)}}{{\dfrac{1}{2} \times 2\cos 75^\circ \cos 15^\circ }}$

We know that $2\cos A\cos B = \cos \left( {A + B} \right) + \cos \left( {A - B} \right)$. Therefore, we get

$\Rightarrow \dfrac{{2\sin \left( {75^\circ - 15^\circ } \right)}}{{\cos \left( {75^\circ + 15^\circ } \right) + \cos \left( {75^\circ - 15^\circ } \right)}}$

$\Rightarrow \dfrac{{2\sin 60^\circ }}{{\cos 90^\circ + \cos 60^\circ }}$

On substituting the values, we get

$\Rightarrow \dfrac{{2 \times \dfrac{{\sqrt 3 }}{2}}}{{0 + \dfrac{1}{2}}}$

$\Rightarrow \dfrac{{\sqrt 3 }}{{0 + \dfrac{1}{2}}} = 2\sqrt 3$

Hence, option (a) is the correct answer.

1.  Which of the following is correct?

1. $\sin 1^\circ > \sin 1$

2. $\sin 1^\circ < \sin 1$

3. $\sin 1^\circ = \sin 1$

4. $\sin 1^\circ = \dfrac{\pi }{{18^\circ }}\sin 1$

Ans: We know that when $\theta$ increases, the value of $\sin \theta$ also increases.

Therefore, $\sin 1^\circ < \sin 1$

(Since $1radian = \dfrac{\pi }{{180}}\sin 1$ )

Hence, option (d) is the correct answer.

1.  If $\tan \alpha = \dfrac{m}{{m + 1}}$, $\tan \beta = \dfrac{1}{{2m + 1}}$, then $\alpha + \beta$ is equal to

1. $\dfrac{\pi }{2}$

2. $\dfrac{\pi }{3}$

3. $\dfrac{\pi }{6}$

4. $\dfrac{\pi }{4}$

Ans: We have, $\tan \alpha = \dfrac{m}{{m + 1}}$ and $\tan \beta = \dfrac{1}{{2m + 1}}$

We know that $\tan (A + B) = \dfrac{{\tan A + \tan B}}{{1 - \tan A.\tan B}}$. Therefore on substituting the values, we get

$\Rightarrow \tan (\alpha + \beta ) = \dfrac{{\dfrac{m}{{m + 1}} + \dfrac{1}{{2m + 1}}}}{{1 - \dfrac{m}{{m + 1}}.\dfrac{1}{{2m + 1}}}}$

$\Rightarrow \tan (\alpha + \beta ) = \dfrac{{\dfrac{{m\left( {2m + 1} \right) + m + 1}}{{\left( {m + 1} \right)\left( {2m + 1} \right)}}}}{{\dfrac{{\left( {m + 1} \right)\left( {2m + 1} \right) - m}}{{\left( {m + 1} \right)\left( {2m + 1} \right)}}}}$

On canceling common terms, we get

$\Rightarrow \tan (\alpha + \beta ) = \dfrac{{2{m^2} + m + m + 1}}{{2{m^2} + m + 2m + 1 - m}}$

On simplification, we get

$\Rightarrow \tan (\alpha + \beta ) = \dfrac{{2{m^2} + 2m + 1}}{{2{m^2} + 2m + 1}} = 1$

We know that $\tan \dfrac{\pi }{4} = 1$. Therefore, we get

$\Rightarrow \tan (\alpha + \beta ) = \tan \dfrac{\pi }{4}$

$\Rightarrow \alpha + \beta = \dfrac{\pi }{4}$

Hence, option (d) is the correct answer.

1.  The minimum value of $3\cos x + 4\sin x + 8$ is

1. $5$

2. $9$

3. $7$

4. $3$

Ans: Given expression, $3\cos x + 4\sin x + 8$

Let $y = 3\cos x + 4\sin x + 8$

Transport $8$ to the LHS

$\Rightarrow y - 8 = 3\cos x + 4\sin x......\left( i \right)$

Now, we will find the minimum value of $y - 8$.

$\Rightarrow y - 8 = - \sqrt {{3^2} + {4^2}}$

$\Rightarrow y - 8 = - \sqrt {9 + 16}$

$\Rightarrow y - 8 = - \sqrt {25}$

$\Rightarrow y - 8 = - 5$

Now, we will find the minimum value of $3\cos x + 4\sin x + 8$.

$\Rightarrow y = - 5 + 8$

$\Rightarrow y = 3$

Therefore, the minimum value of $3\cos x + 4\sin x + 8$ is $3$.

Hence, option (d) is the correct answer.

1.  The value of $\tan 3A - \tan 2A - \tan A$ is equal to

1. $\tan 3A\tan 2A\tan A$

2. $- \tan 3A\tan 2A\tan A$

3. $\tan A\tan 2A - \tan 2A\tan 3A - \tan 3A\tan A$

4. None of these

Ans: Given expression, $\tan 3A - \tan 2A - \tan A$

We can write $\tan 3A = \tan \left( {2A + A} \right)$.

We know that $\tan \left( {x + y} \right) = \dfrac{{\tan x + \tan y}}{{1 - \tan x\tan y}}$. Therefore, we get

$\Rightarrow \tan 3A = \dfrac{{\tan 2A + \tan A}}{{1 - \tan 2A\tan A}}$

Now, we will cross multiply the above written equation.

$\Rightarrow \tan 3A\left( {1 - \tan 2A\tan A} \right) = \tan 2A + \tan A$

On multiplication of terms, we get

$\Rightarrow \tan 3A - \tan 3A\tan 2A\tan A = \tan 2A + \tan A$

$\Rightarrow \tan 3A - \tan 2A - \tan A = \tan 3A\tan 2A\tan A$

Hence, option (a) is the correct answer.

1.  The value of $\sin \left( {45^\circ + \theta } \right) - \cos \left( {45^\circ - \theta } \right)$ is

1. $2\cos \theta$

2. $2\sin \theta$

3. $1$

4. $0$

Ans: Given, $\sin \left( {45^\circ + \theta } \right) - \cos \left( {45^\circ - \theta } \right)$

As we know $\sin \left( {A + B} \right) = \sin A\cos B + \cos A\sin B$ and $\cos \left( {A - B} \right) = \cos A\cos B + \sin A\sin B$. Therefore, we get

$\Rightarrow \left( {\sin 45^\circ \cos \theta + \cos 45^\circ \sin \theta } \right) - \left( {\cos 45^\circ \cos \theta + \sin 45^\circ \sin \theta } \right)$

Substitute value of $\sin 45^\circ = \dfrac{1}{{\sqrt 2 }}$ and $\cos 45^\circ = \dfrac{1}{{\sqrt 2 }}$.

$\Rightarrow \left( {\dfrac{1}{{\sqrt 2 }}\cos \theta + \dfrac{1}{{\sqrt 2 }}\sin \theta } \right) - \left( {\dfrac{1}{{\sqrt 2 }}\cos \theta + \dfrac{1}{{\sqrt 2 }}\sin \theta } \right)$

$\Rightarrow \dfrac{1}{{\sqrt 2 }}\cos \theta + \dfrac{1}{{\sqrt 2 }}\sin \theta - \dfrac{1}{{\sqrt 2 }}\cos \theta - \dfrac{1}{{\sqrt 2 }}\sin \theta$

$\Rightarrow 0$

Hence, option (d) is the correct answer.

1.  The value of $\cot \left( {\dfrac{\pi }{4} + \theta } \right)\cot \left( {\dfrac{\pi }{4} - \theta } \right)$ is

1. $- 1$

2. $0$

3. $1$

4. Not defined

Ans: We have, $\cot \left( {\dfrac{\pi }{4} + \theta } \right)\cot \left( {\dfrac{\pi }{4} - \theta } \right)$

We know that $\cot \left( {x + y} \right) = \dfrac{{\cot x.\cot y - 1}}{{\cot y + \cot x}}$ and $\cot \left( {x - y} \right) = \dfrac{{\cot x.\cot y + 1}}{{\cot y - \cot x}}$. Therefore, we get

$\Rightarrow \dfrac{{\cot \dfrac{\pi }{4}.\cot \theta - 1}}{{\cot \theta + \cot \dfrac{\pi }{4}}} \times \dfrac{{\cot \dfrac{\pi }{4}.\cot \theta + 1}}{{\cot \theta - \cot \dfrac{\pi }{4}}}$

Substitute value of $\cos \dfrac{\pi }{4} = 1$.

$\Rightarrow \dfrac{{1.\cot \theta - 1}}{{\cot \theta + 1}} \times \dfrac{{1.\cot \theta + 1}}{{\cot \theta - 1}}$

$\Rightarrow \dfrac{{\cot \theta - 1}}{{\cot \theta + 1}} \times \dfrac{{\cot \theta + 1}}{{\cot \theta - 1}}$

On canceling common terms, we get

$\Rightarrow 1$

Hence, option (c) is the correct answer.

1. $\cos 2\theta \cos 2\phi + {\sin ^2}\left( {\theta - \phi } \right) - {\sin ^2}\left( {\theta + \phi } \right)$ is equal to

1. $\sin 2\left( {\theta + \phi } \right)$

2. $\cos 2\left( {\theta + \phi } \right)$

3. $\sin 2\left( {\theta - \phi } \right)$

4. $\cos 2\left( {\theta - \phi } \right)$

Ans: We have, $\cos 2\theta \cos 2\phi + {\sin ^2}\left( {\theta - \phi } \right) - {\sin ^2}\left( {\theta + \phi } \right)$

We know that ${\sin ^2}A - {\sin ^2}B = \sin \left( {A + B} \right).\sin \left( {A - B} \right)$. Therefore, we get

$\Rightarrow \cos 2\theta \cos 2\phi + \sin \left( {\theta - \phi + \theta + \phi } \right).\sin \left( {\theta - \phi - \theta - \phi } \right)$

$\Rightarrow \cos 2\theta \cos 2\phi + \sin \left( {2\theta } \right).\sin \left( { - 2\phi } \right)$

We know that $\sin \left( { - \theta } \right) = - \sin \theta$. Therefore, we get

$\Rightarrow \cos 2\theta \cos 2\phi - \sin 2\theta \sin 2\phi$

We know that $\cos \left( {A + B} \right) = \cos A\cos B - \sin A\sin B$. Therefore, we get

$\Rightarrow \cos 2\left( {\theta + \phi } \right)$

Hence, option (b) is the correct answer.

1.  The value of $\cos 12^\circ + \cos 84^\circ + \cos 156^\circ + \cos 132^\circ$ is

1. $\dfrac{1}{2}$

2. $1$

3. $- \dfrac{1}{2}$

4. $\dfrac{1}{8}$

Ans: Given expression: $\cos 12^\circ + \cos 84^\circ + \cos 156^\circ + \cos 132^\circ$

$\Rightarrow \left( {\cos 132^\circ + \cos 12^\circ } \right) + \left( {\cos 156^\circ + \cos 84^\circ } \right)$

We know that $\cos A + \cos B = 2\cos \dfrac{{A + B}}{2}\cos \dfrac{{A - B}}{2}$. Therefore, we get

$\Rightarrow 2\cos \left( {\dfrac{{132^\circ + 12^\circ }}{2}} \right).\cos \left( {\dfrac{{132^\circ - 12^\circ }}{2}} \right) + 2\cos \left( {\dfrac{{156^\circ + 84^\circ }}{2}} \right).\cos \left( {\dfrac{{156^\circ - 84^\circ }}{2}} \right)$

$\Rightarrow 2\cos \left( {\dfrac{{144^\circ }}{2}} \right).\cos \left( {\dfrac{{120^\circ }}{2}} \right) + 2\cos \left( {\dfrac{{240^\circ }}{2}} \right).\cos \left( {\dfrac{{72^\circ }}{2}} \right)$

$\Rightarrow 2\cos \left( {72^\circ } \right).\cos \left( {60^\circ } \right) + 2\cos \left( {120^\circ } \right).\cos \left( {36^\circ } \right)$

Substitute value of $\cos 60^\circ = \dfrac{1}{2}$ and $\cos 120^\circ = - \dfrac{1}{2}$.

$\Rightarrow 2\cos \left( {72^\circ } \right) \times \dfrac{1}{2} \times + 2 \times - \dfrac{1}{2} \times \cos \left( {36^\circ } \right)$

$\Rightarrow \cos 72^\circ - \cos 36^\circ$

$\Rightarrow \cos \left( {90^\circ - 18^\circ } \right) - \cos 36^\circ$

We know that $\cos \left( {90^\circ - \theta } \right) = \sin \theta$. Therefore, we get

$\Rightarrow \sin 18^\circ - \cos 36^\circ$

Substitute value of $\sin 18^\circ = \dfrac{{\sqrt 5 - 1}}{4}$ and $\cos 36^\circ = \dfrac{{\sqrt 5 + 1}}{4}$.

$\Rightarrow \dfrac{{\sqrt 5 - 1}}{4} - \dfrac{{\sqrt 5 + 1}}{4}$

$\Rightarrow \dfrac{{\sqrt 5 - 1 - \sqrt 5 - 1}}{4}$

$\Rightarrow \dfrac{{ - 2}}{4}$

$\Rightarrow \dfrac{{ - 1}}{2}$

Hence, option (c) is the correct answer.

1.  If $\tan A = \dfrac{1}{2}$, $\tan B = \dfrac{1}{3}$, then $\tan \left( {2A + B} \right)$ is equal to

1. $1$

2. $2$

3. $3$

4. $4$

Ans: Given, $\tan A = \dfrac{1}{2}$, $\tan B = \dfrac{1}{3}$

We know that $\tan 2x = \dfrac{{2\tan x}}{{1 - {{\tan }^2}x}}$. Therefore, we get

$\Rightarrow \tan 2A = \dfrac{{2\tan A}}{{1 - {{\tan }^2}A}}$

Now, we will substitute the value of $\tan A = \dfrac{1}{2}$

$\Rightarrow \tan 2A = \dfrac{{2 \times \dfrac{1}{2}}}{{1 - \dfrac{{{1^2}}}{{{2^2}}}}}$

$\Rightarrow \tan 2A = \dfrac{1}{{\dfrac{{4 - 1}}{4}}} = \dfrac{1}{{\dfrac{3}{4}}}$

$\Rightarrow \tan 2A = \dfrac{4}{3}$

We need to find value of $\tan \left( {2A + B} \right)$

We know that $\tan \left( {x + y} \right) = \dfrac{{\tan x + \tan y}}{{1 - \tan x.\tan y}}$. Therefore, we get

$\Rightarrow \tan \left( {2A + B} \right) = \dfrac{{\tan 2A + \tan B}}{{1 - \tan 2A.\tan B}}$

$\Rightarrow \tan \left( {2A + B} \right) = \dfrac{{\dfrac{4}{3} + \dfrac{1}{3}}}{{1 - \dfrac{4}{3}.\dfrac{1}{3}}}$

$\Rightarrow \tan \left( {2A + B} \right) = \dfrac{{\dfrac{5}{3}}}{{\dfrac{{9 - 4}}{9}}}$

$\Rightarrow \tan \left( {2A + B} \right) = \dfrac{5}{3} \times \dfrac{9}{5}$

$\Rightarrow \tan \left( {2A + B} \right) = 3$

Hence, option (c) is the correct answer.

1.  The value of $\sin \dfrac{\pi }{{10}}\sin \dfrac{{13\pi }}{{10}}$ is

1. $\dfrac{1}{2}$

2. $- \dfrac{1}{2}$

3. $- \dfrac{1}{4}$

4. $1$

Ans: We have, $\sin \dfrac{\pi }{{10}}\sin \dfrac{{13\pi }}{{10}}$

$\Rightarrow \sin \dfrac{\pi }{{10}}\sin \left( {\pi + \dfrac{{3\pi }}{{10}}} \right)$

We know that $\sin \left( {\pi + \theta } \right) = - \sin \theta$. Therefore, we get

$\Rightarrow \sin \dfrac{\pi }{{10}}\sin \left( { - \dfrac{{3\pi }}{{10}}} \right)$

We know that $\sin \left( { - \theta } \right) = - \sin \theta$. Therefore, we get

$\Rightarrow - \sin \dfrac{{180^\circ }}{{10}}\sin \left( {\dfrac{{3 \times 180^\circ }}{{10}}} \right)$

$\Rightarrow - \sin 18^\circ .\sin 54^\circ$

$\Rightarrow - \sin 18^\circ .\sin \left( {90^\circ - 36^\circ } \right)$

We know that $\sin \left( {90^\circ - \theta } \right) = \cos \theta$. Therefore, we get

$\Rightarrow - \sin 18^\circ .\cos 36^\circ$

On substituting the values $\sin 18^\circ = \dfrac{{\sqrt 5 - 1}}{4}$ and $\cos 36^\circ = \dfrac{{\sqrt 5 + 1}}{4}$, we get

$\Rightarrow - \left( {\dfrac{{\sqrt 5 - 1}}{4}} \right)\left( {\dfrac{{\sqrt 5 + 1}}{4}} \right)$

$\Rightarrow - \left( {\dfrac{{5 - 1}}{{16}}} \right)$

$\Rightarrow - \left( {\dfrac{4}{{16}}} \right) = - \dfrac{1}{4}$

Hence, option (c) is the correct answer.

1.  The value of $\sin 50^\circ - \sin 70^\circ + \sin 10^\circ$ is equal to

1. $1$

2. $0$

3. $\dfrac{1}{2}$

4. $2$

Ans: Given expression, $\sin 50^\circ - \sin 70^\circ + \sin 10^\circ$

$\Rightarrow \left( {\sin 50^\circ - \sin 70^\circ } \right) + \sin 10^\circ$

We know that $\sin C - \sin D = 2\cos \dfrac{{C + D}}{2}\sin \dfrac{{C - D}}{2}$. Therefore, we get

$\Rightarrow 2\cos \dfrac{{50^\circ + 70^\circ }}{2}\sin \dfrac{{50^\circ - 70^\circ }}{2} + \sin 10^\circ$

$\Rightarrow 2\cos 60^\circ \sin \left( { - 10^\circ } \right) + \sin 10^\circ$

We know that $\sin \left( { - \theta } \right) = - \sin \theta$. Therefore, we get

$\Rightarrow - 2\cos 60^\circ \sin 10^\circ + \sin 10^\circ$

Substitute value of $\sin 60^\circ = \dfrac{1}{2}$.

$\Rightarrow - 2 \times \dfrac{1}{2} \times \sin 10^\circ + \sin 10^\circ$

$\Rightarrow - \sin 10^\circ + \sin 10^\circ$

$\Rightarrow 0$

Hence, option (b) is the correct answer.

1.  If $\sin \theta + \cos \theta = 1$, then the value of $\sin 2\theta$ is equal to

1. $1$

2. $\dfrac{1}{2}$

3. $0$

4. $- 1$

Ans: Given, $\sin \theta + \cos \theta = 1$

On squaring both the sides, we get

$\Rightarrow {\left( {\sin \theta + \cos \theta } \right)^2} = {1^2}$

As we know ${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$. So, on applying this identity, we get

$\Rightarrow {\sin ^2}\theta + {\cos ^2}\theta + 2\sin \theta \cos \theta = 1$

As we know ${\sin ^2}\theta + {\cos ^2}\theta = 1$. So, on applying this formula, we get

$\Rightarrow 1 + 2\sin \theta \cos \theta = 1$

As we know $\sin 2A = 2\sin A\cos A$. Therefore, we get

$\Rightarrow 2\sin \theta \cos \theta = 1 - 1$

$\Rightarrow \sin 2\theta = 0$

Hence, option (c) is the correct answer.

1.  If $\alpha + \beta = \dfrac{\pi }{4}$, then the value of $\left( {1 + \tan \alpha } \right)\left( {1 + \tan \beta } \right)$ is

1. $1$

2. $2$

3. $- 2$

4. Not defined

Ans: Given, $\alpha + \beta = \dfrac{\pi }{4}$

$\Rightarrow \tan \left( {\alpha + \beta } \right) = \tan \dfrac{\pi }{4}$

We know that $\tan \left( {x + y} \right) = \dfrac{{\tan x + \tan y}}{{1 - \tan x\tan y}}$. Therefore, we get

$\Rightarrow \dfrac{{\tan \alpha + \tan \beta }}{{1 - \tan \alpha \tan \beta }} = \tan \dfrac{\pi }{4}$

On substituting the value of $\tan \dfrac{\pi }{4} = 1$, we get

$\Rightarrow \dfrac{{\tan \alpha + \tan \beta }}{{1 - \tan \alpha \tan \beta }} = 1$

On cross-multiplication, we get

$\Rightarrow \tan \alpha + \tan \beta = 1 - \tan \alpha \tan \beta$

$\Rightarrow \tan \alpha + \tan \beta + \tan \alpha \tan \beta = 1$

Add $1$ to both the sides

$\Rightarrow 1 + \tan \alpha + \tan \beta + \tan \alpha \tan \beta = 1 + 1$

Take $\tan \beta$ as a common term.

$\Rightarrow \left( {1 + \tan \alpha } \right) + \tan \beta \left( {1 + \tan \alpha } \right) = 2$

$\Rightarrow \left( {1 + \tan \alpha } \right)\left( {1 + \tan \alpha } \right) = 2$

Hence, option (b) is the correct answer.

1.  If $\sin \theta = \dfrac{{ - 4}}{5}$ and $\theta$ lies in third quadrant then the value of $\cos \dfrac{\theta }{2}$ is

1. $\dfrac{1}{5}$

2. $- \dfrac{1}{{\sqrt {10} }}$

3. $- \dfrac{1}{{\sqrt 5 }}$

4. $\dfrac{1}{{\sqrt {10} }}$

Ans: Given, $\sin \theta = \dfrac{{ - 4}}{5}$. Here, value of $\sin e$ is negative because $\theta$ lies in third quadrant. As here we need to find $\cos \dfrac{\theta }{2}$, so we will first find the value of $\cos \theta$.

We know that ${\sin ^2}\theta + {\cos ^2}\theta = 1$.

$\Rightarrow \cos \theta = \sqrt {1 - {{\sin }^2}\theta }$

Substitute the value of $\sin \theta = \dfrac{{ - 4}}{5}$.

$\Rightarrow \cos \theta = \sqrt {1 - {{\left( {\dfrac{{ - 4}}{5}} \right)}^2}}$

$\Rightarrow \cos \theta = \sqrt {1 - \dfrac{{16}}{{25}}}$

On taking LCM, we get

$\Rightarrow \cos \theta = \sqrt {\dfrac{{25 - 16}}{{25}}}$

$\Rightarrow \cos \theta = \sqrt {\dfrac{9}{{25}}}$

$\Rightarrow \cos \theta = \pm \dfrac{3}{5}$

As it is mentioned in the question that $\theta$ lies in the third quadrant and we know $\cos$ is negative in the third quadrant. Therefore, we get

$\Rightarrow \cos \theta = - \dfrac{3}{5}$

We know that $\cos \theta = 2{\cos ^2}\dfrac{\theta }{2} - 1$. Now we will substitute value of $\cos \theta$ in this identity.

$\Rightarrow \dfrac{{ - 3}}{5} = 2{\cos ^2}\dfrac{\theta }{2} - 1$

(Here, value of $\cos \theta$ is negative because $\pi < \theta < \dfrac{{3\pi }}{2}$ )

$\Rightarrow 2{\cos ^2}\dfrac{\theta }{2} = 1 - \dfrac{3}{5}$

$\Rightarrow 2{\cos ^2}\dfrac{\theta }{2} = \dfrac{2}{5}$

On canceling common term, we get

$\Rightarrow {\cos ^2}\dfrac{\theta }{2} = \dfrac{1}{5}$

$\Rightarrow \cos \dfrac{\theta }{2} = \pm \dfrac{1}{{\sqrt 5 }}$

As we know $\pi < \theta < \dfrac{{3\pi }}{2}$. Therefore, $\dfrac{\pi }{2} < \dfrac{\theta }{2} < \dfrac{{3\pi }}{4}$

$\Rightarrow \cos \dfrac{\theta }{2} = - \dfrac{1}{{\sqrt 5 }}$

Hence, option (c) is the correct answer.

1.  Number of solutions of the equation $\tan x + \sec x = 2\cos x$ lying in the interval $\left[ {0,2\pi } \right]$ is

1. $0$

2. $1$

3. $2$

4. $3$

Ans: Given that, $\tan x + \sec x = 2\cos x$

We will first write above written equation in terms of $\sin e$ and $\cos$.

$\Rightarrow \dfrac{{\sin x}}{{\cos x}} + \dfrac{1}{{\cos x}} = 2\cos x$

$\Rightarrow \dfrac{{\sin x + 1}}{{\cos x}} = 2\cos x$

$\Rightarrow \sin x + 1 = 2{\cos ^2}x$

$\Rightarrow 2{\cos ^2}x - \sin x - 1 = 0$

$\Rightarrow 2\left( {1 - {{\sin }^2}x} \right) - \sin x - 1 = 0$

$\Rightarrow - 2{\sin ^2}x - \sin x + 1 = 0$

$\Rightarrow 2{\sin ^2}x + \sin x - 1 = 0$

Since, the above written equation is a quadratic equation in $\sin x$. Therefore, it will have $2$ solutions.

Hence, option (c) is the correct answer.

1.  The value of $\sin \dfrac{\pi }{{18}} + \sin \dfrac{\pi }{9} + \sin \dfrac{{2\pi }}{9} + \sin \dfrac{{5\pi }}{{18}}$ is given by

1. $\sin \dfrac{{7\pi }}{{18}} + \sin \dfrac{{4\pi }}{9}$

2. $1$

3. $\cos \dfrac{\pi }{6} + \cos \dfrac{{3\pi }}{7}$

4. $\cos \dfrac{\pi }{9} + \sin \dfrac{\pi }{9}$

Ans: Given expression: $\sin \dfrac{\pi }{{18}} + \sin \dfrac{\pi }{9} + \sin \dfrac{{2\pi }}{9} + \sin \dfrac{{5\pi }}{{18}}$

$\Rightarrow \left( {\sin \dfrac{{5\pi }}{{18}} + \sin \dfrac{\pi }{{18}}} \right) + \left( {\sin \dfrac{{2\pi }}{9} + \sin \dfrac{\pi }{9}} \right)$

We know that $\sin A + \sin B = 2\sin \dfrac{{A + B}}{2}\cos \dfrac{{A - B}}{2}$. Therefore, we get

$\Rightarrow 2\sin \left( {\dfrac{{\dfrac{{5\pi }}{{18}} + \dfrac{\pi }{{18}}}}{2}} \right).\cos \left( {\dfrac{{\dfrac{{5\pi }}{{18}} - \dfrac{\pi }{{18}}}}{2}} \right) + 2\sin \left( {\dfrac{{\dfrac{{2\pi }}{9} + \dfrac{\pi }{9}}}{2}} \right).\cos \left( {\dfrac{{\dfrac{{2\pi }}{9} - \dfrac{\pi }{9}}}{2}} \right)$

$\Rightarrow 2\sin \left( {\dfrac{{\dfrac{{6\pi }}{{18}}}}{2}} \right).\cos \left( {\dfrac{{\dfrac{{4\pi }}{{18}}}}{2}} \right) + 2\sin \left( {\dfrac{{\dfrac{{3\pi }}{9}}}{2}} \right).\cos \left( {\dfrac{{\dfrac{\pi }{9}}}{2}} \right)$

$\Rightarrow 2\sin \left( {\dfrac{\pi }{6}} \right).\cos \left( {\dfrac{\pi }{9}} \right) + 2\sin \left( {\dfrac{\pi }{6}} \right).\cos \left( {\dfrac{\pi }{{18}}} \right)$

Substitute value of $\sin \dfrac{\pi }{6} = \dfrac{1}{2}$.

$\Rightarrow 2 \times \dfrac{1}{2} \times \cos \left( {\dfrac{\pi }{9}} \right) + 2 \times \dfrac{1}{2} \times \cos \left( {\dfrac{\pi }{{18}}} \right)$

$\Rightarrow \cos \left( {\dfrac{\pi }{9}} \right) + \cos \left( {\dfrac{\pi }{{18}}} \right)$

We know that $\cos \theta = \sin \left( {90^\circ - \theta } \right)$. Therefore, we get

$\Rightarrow \sin \left( {\dfrac{\pi }{2} - \dfrac{\pi }{9}} \right) + \sin \left( {\dfrac{\pi }{2} - \dfrac{\pi }{{18}}} \right)$

$\Rightarrow \sin \dfrac{{7\pi }}{{18}} + \sin \dfrac{{8\pi }}{{18}}$

$\Rightarrow \sin \dfrac{{7\pi }}{{18}} + \sin \dfrac{{4\pi }}{9}$

Hence, option (a) is the correct answer.

1.  If $A$ lies in the second quadrant and $3\tan A + 4 = 0$, then the value of $2\cot A - 5\cos A + \sin A$ is equal to

1. $\dfrac{{ - 53}}{{10}}$

2. $\dfrac{{23}}{{10}}$

3. $\dfrac{{37}}{{10}}$

4. $\dfrac{7}{{10}}$

Ans: Given, $3\tan A + 4 = 0$

$\Rightarrow 3\tan A = - 4$

$\Rightarrow \tan A = \dfrac{{ - 4}}{3} = \dfrac{P}{B}$

We know that ${H^2} = {P^2} + {B^2}$

$\Rightarrow {H^2} = {\left( { - 4} \right)^2} + {\left( 3 \right)^2}$

$\Rightarrow {H^2} = 16 + 9$

$\Rightarrow {H^2} = 25$

$\Rightarrow H = 5$

We know that $\cos x = \dfrac{B}{H}$. Therefore, we get

$\Rightarrow \cos A = \dfrac{{ - 3}}{5}$ (Negative because $A$ lies in second quadrant)

We know that $\sin x = \dfrac{P}{H}$. Therefore, we get

$\Rightarrow \sin A = \dfrac{4}{5}$

We know that $\cot x = \dfrac{B}{P}$. Therefore, we get

$\Rightarrow \cot A = \dfrac{{ - 3}}{4}$ (Negative because $A$ lies in second quadrant)

We need to find the value of $2\cot A - 5\cos A + \sin A$

$\Rightarrow 2 \times \dfrac{{ - 3}}{4} - 5 \times \left( {\dfrac{{ - 3}}{5}} \right) + \dfrac{4}{5}$

$\Rightarrow \dfrac{{ - 3}}{2} + 3 + \dfrac{4}{5}$

$\Rightarrow \dfrac{{ - 15 + 30 + 8}}{{10}}$

$\Rightarrow \dfrac{{23}}{{10}}$

Hence, option (a) is the correct answer.

1.  The value of ${\cos ^2}48^\circ - {\sin ^2}12^\circ$ is

1. $\dfrac{{\sqrt 5 + 1}}{8}$

2. $\dfrac{{\sqrt 5 - 1}}{8}$

3. $\dfrac{{\sqrt 5 + 1}}{5}$

4. $\dfrac{{\sqrt 5 + 1}}{{2\sqrt 2 }}$

Ans: Given expression: ${\cos ^2}48^\circ - {\sin ^2}12^\circ$

We know that ${\cos ^2}A - {\sin ^2}A = \cos \left( {A + B} \right).\cos \left( {A - B} \right)$. Therefore, we get

$\Rightarrow \cos \left( {48^\circ + 12^\circ } \right).\cos \left( {48^\circ - 12^\circ } \right)$

$\Rightarrow \cos 60^\circ .\cos 36^\circ$

On substituting the values, $\cos 60^\circ = \dfrac{1}{2}$ and $\cos 36^\circ = \dfrac{{\sqrt 5 + 1}}{4}$. Therefore, we get

$\Rightarrow \dfrac{1}{2} \times \dfrac{{\sqrt 5 + 1}}{4}$

$\Rightarrow \dfrac{{\sqrt 5 + 1}}{8}$

Hence, option (a) is the correct answer.

1.  If $\tan \alpha = \dfrac{1}{7}$, $\tan \beta = \dfrac{1}{3}$, then $\cos 2\alpha$ is equal to

1. $\sin 2\beta$

2. $\sin 4\beta$

3. $\sin 3\beta$

4. $\cos 2\beta$

Ans: Given, $\tan \alpha = \dfrac{1}{7}$, $\tan \beta = \dfrac{1}{3}$

We know that $\cos 2x = \dfrac{{1 - {{\tan }^2}x}}{{1 + {{\tan }^2}x}}$. Therefore, we get

$\Rightarrow \cos 2\alpha = \dfrac{{1 - {{\tan }^2}\alpha }}{{1 + {{\tan }^2}\alpha }} = \dfrac{{1 - \dfrac{{{1^2}}}{{{7^2}}}}}{{1 + \dfrac{{{1^2}}}{{{7^2}}}}}$

$\Rightarrow \cos 2\alpha = \dfrac{{\dfrac{{{7^2} - 1}}{{{7^2}}}}}{{\dfrac{{{7^2} + 1}}{{{7^2}}}}} = \dfrac{{{7^2} - 1}}{{{7^2} + 1}}$

$\Rightarrow \cos 2\alpha = \dfrac{{49 - 1}}{{49 + 1}} = \dfrac{{48}}{{50}}$

$\Rightarrow \cos 2\alpha = \dfrac{{24}}{{25}}$

Similarly,

$\Rightarrow \tan 2\beta = \dfrac{{2\tan \beta }}{{1 - {{\tan }^2}\beta }}$

$\Rightarrow \tan 2\beta = \dfrac{{2 \times \dfrac{1}{3}}}{{1 - \dfrac{{{1^2}}}{{{3^2}}}}}$

$\Rightarrow \tan 2\beta = \dfrac{{\dfrac{2}{3}}}{{\dfrac{8}{9}}} = \dfrac{2}{3} \times \dfrac{9}{8}$

$\Rightarrow \tan 2\beta = \dfrac{3}{4}$

We know that $\sin 2x = \dfrac{{2\tan x}}{{1 + {{\tan }^2}x}}$. Therefore, we get

$\Rightarrow \sin 4\beta = \dfrac{{2\tan 2\beta }}{{1 + {{\tan }^2}2\beta }}$

$\Rightarrow \sin 4\beta = \dfrac{{2 \times \dfrac{3}{4}}}{{1 + \dfrac{{{3^2}}}{{{4^2}}}}}$

$\Rightarrow \sin 4\beta = \dfrac{{\dfrac{6}{4}}}{{\dfrac{{16 + 9}}{{16}}}}$

$\Rightarrow \sin 4\beta = \dfrac{6}{4} \times \dfrac{{16}}{{25}}$

$\Rightarrow \sin 4\beta = \dfrac{{24}}{{25}} = \cos 2\alpha$

Hence, option (b) is the correct answer.

1.  If $\tan \theta = \dfrac{a}{b}$, then $b\cos 2\theta + a\sin 2\theta$ is equal to

1. $a$

2. $b$

3. $\dfrac{a}{b}$

4. None

Ans: Given, $\tan \theta = \dfrac{a}{b}$

We have, $b\cos 2\theta + a\sin 2\theta$

We know that $\cos 2x = \dfrac{{1 - {{\tan }^2}x}}{{1 + {{\tan }^2}x}}$ and $\sin 2x = \dfrac{{2\tan x}}{{1 + {{\tan }^2}x}}$. Therefore, we get

$\Rightarrow b\left[ {\dfrac{{1 - {{\tan }^2}\theta }}{{1 + {{\tan }^2}\theta }}} \right] + a\left[ {\dfrac{{2\tan \theta }}{{1 + {{\tan }^2}\theta }}} \right]$

Substitute $\tan \theta = \dfrac{a}{b}$

$\Rightarrow b\left[ {\dfrac{{1 - \dfrac{{{a^2}}}{{{b^2}}}}}{{1 + \dfrac{{{a^2}}}{{{b^2}}}}}} \right] + a\left[ {\dfrac{{2\dfrac{a}{b}}}{{1 + \dfrac{{{a^2}}}{{{b^2}}}}}} \right]$

$\Rightarrow b\left[ {\dfrac{{\dfrac{{{b^2} - {a^2}}}{{{b^2}}}}}{{\dfrac{{{b^2} + {a^2}}}{{{b^2}}}}}} \right] + a\left[ {\dfrac{{2\dfrac{a}{b}}}{{\dfrac{{{b^2} + {a^2}}}{{{b^2}}}}}} \right]$

$\Rightarrow b\left[ {\dfrac{{{b^2} - {a^2}}}{{{b^2} + {a^2}}}} \right] + a\left[ {\dfrac{{2ab}}{{{b^2} + {a^2}}}} \right]$

$\Rightarrow \left[ {\dfrac{{{b^3} - {a^2}b}}{{{b^2} + {a^2}}}} \right] + \left[ {\dfrac{{2{a^2}b}}{{{b^2} + {a^2}}}} \right]$

$\Rightarrow \dfrac{{{b^3} - {a^2}b + 2{a^2}b}}{{{b^2} + {a^2}}}$

$\Rightarrow \dfrac{{b\left( {{b^2} + {a^2}} \right)}}{{{b^2} + {a^2}}}$

$\Rightarrow b$

Hence, option (b) is the correct answer.

1.  If for real values of $x$, $\cos \theta = x + \dfrac{1}{x}$, then

1. $\theta$ is an acute angle

2. $\theta$ is right angle

3. $\theta$ is an obtuse angle

4. No value of $\theta$ is possible

Ans: Given, $\cos \theta = x + \dfrac{1}{x}$

$\Rightarrow \cos \theta = \dfrac{{{x^2} + 1}}{x}$

$\Rightarrow {x^2} + 1 = x\cos \theta$

$\Rightarrow {x^2} - x\cos \theta + 1 = 0$

We know that for real value of $x$, ${b^2} - 4ac \geqslant 0$. Therefore, we get

$\Rightarrow {\left( { - \cos \theta } \right)^2} - 4 \times 1 \times 1 \geqslant 0$

$\Rightarrow {\left( { - \cos \theta } \right)^2} - 4 \geqslant 0$

$\Rightarrow {\cos ^2}\theta \geqslant 4$

$\Rightarrow \cos \theta \geqslant \pm 2$

We know that $- 1 \leqslant \cos \theta \leqslant 1$. Therefore, value of $\theta$ is not possible

Hence, option (d) is the correct answer.

Fill in the blanks in exercises $60$ to $67$.

1.  The value of $\dfrac{{\sin 50^\circ }}{{\sin 130^\circ }}$ is …….

Ans: We need to find the value of $\dfrac{{\sin 50^\circ }}{{\sin 130^\circ }}$

As we know $\sin \left( {180^\circ - \theta } \right) = \sin \theta$. Therefore, we get

$\Rightarrow \dfrac{{\sin 50^\circ }}{{\sin \left( {180^\circ - 50^\circ } \right)}}$

$\Rightarrow \dfrac{{\sin 50^\circ }}{{\sin 50^\circ }}$

On canceling the common term, we get

$\Rightarrow 1$

Thus, value of filler is $1$.

1.  If $k = \sin \left( {\dfrac{\pi }{{18}}} \right)\sin \left( {\dfrac{{5\pi }}{{18}}} \right)\sin \left( {\dfrac{{7\pi }}{{18}}} \right)$, then the numerical value of $k$ is …….

Ans: Given, $k = \sin \left( {\dfrac{\pi }{{18}}} \right)\sin \left( {\dfrac{{5\pi }}{{18}}} \right)\sin \left( {\dfrac{{7\pi }}{{18}}} \right)$

Substitute value of $\pi = 180^\circ$.

$\Rightarrow k = \sin \left( {\dfrac{{180^\circ }}{{18}}} \right)\sin \left( {\dfrac{{5 \times 180^\circ }}{{18}}} \right)\sin \left( {\dfrac{{7 \times 180^\circ }}{{18}}} \right)$

On simplification, we get

$\Rightarrow k = \sin 10^\circ .\sin 50^\circ .\sin 70^\circ$

$\Rightarrow k = \sin 10^\circ .\sin \left( {90^\circ - 40^\circ } \right).\sin \left( {90^\circ - 20^\circ } \right)$

As we know $\sin \left( {90^\circ - \theta } \right) = \cos \theta$. Therefore, we get

$\Rightarrow k = \sin 10^\circ .\cos 40^\circ .\cos 20^\circ$

Multiply and divide above written equation by $2$

$\Rightarrow k = \sin 10^\circ \dfrac{1}{2}\left[ {2\cos 40^\circ \cos 20^\circ } \right]$

We know that $2\cos x\cos y = \cos \left( {x + y} \right) + \cos \left( {x - y} \right)$. Therefore, we get

$\Rightarrow k = \dfrac{1}{2}\sin 10^\circ \left[ {\cos \left( {40^\circ + 20^\circ } \right) + \cos \left( {40^\circ - 20^\circ } \right)} \right]$

$\Rightarrow k = \dfrac{1}{2}\sin 10^\circ \left[ {\cos 60^\circ + \cos 20^\circ } \right]$

Substitute value of $\cos 60^\circ = \dfrac{1}{2}$.

$\Rightarrow k = \dfrac{1}{2}\sin 10^\circ \left[ {\dfrac{1}{2} + \cos 20^\circ } \right]$

On multiplication of terms, we get

$\Rightarrow k = \dfrac{1}{4}\sin 10^\circ + \dfrac{1}{2}\sin 10^\circ \cos 20^\circ$

Multiply and divide $\sin 10^\circ \cos 20^\circ$ by $2$

$\Rightarrow k = \dfrac{1}{4}\sin 10^\circ + \dfrac{1}{2} \times \dfrac{1}{2}\left( {2\sin 10^\circ \cos 20^\circ } \right)$

We know that $2\sin x\cos y = \sin \left( {x + y} \right) + \sin \left( {x - y} \right)$. Therefore, we get

$\Rightarrow k = \dfrac{1}{4}\sin 10^\circ + \dfrac{1}{4}\left( {\sin \left( {10^\circ + 20^\circ } \right) + \sin \left( {10^\circ - 20^\circ } \right)} \right)$

$\Rightarrow k = \dfrac{1}{4}\sin 10^\circ + \dfrac{1}{4}\left( {\sin 30^\circ + \sin \left( { - 10^\circ } \right)} \right)$

We know that $\sin \left( { - \theta } \right) = - \sin \theta$ . Therefore, we get

$\Rightarrow k = \dfrac{1}{4}\sin 10^\circ + \dfrac{1}{4}\sin 30^\circ - \dfrac{1}{4}\sin 10^\circ$

$\Rightarrow k = \dfrac{1}{4}\sin 30^\circ$

Substitute the value of $\sin 30^\circ = \dfrac{1}{2}$.

$\Rightarrow k = \dfrac{1}{4} \times \dfrac{1}{2} = \dfrac{1}{8}$

Thus, value of filler is $\dfrac{1}{8}$.

1.  If $\tan A = \dfrac{{1 - \cos B}}{{\sin B}}$, then $\tan 2A =$ …….

Ans: Given, $\tan A = \dfrac{{1 - \cos B}}{{\sin B}}$

We know that $\tan 2A = \dfrac{{2\tan A}}{{1 - {{\tan }^2}A}}$. So, we will substitute given value of $\tan A$ in this formula.

$\Rightarrow \tan 2A = \dfrac{{2\left( {\dfrac{{1 - \cos B}}{{\sin B}}} \right)}}{{1 - {{\left( {\dfrac{{1 - \cos B}}{{\sin B}}} \right)}^2}}}$

We know that $1 - \cos x = 2{\sin ^2}\dfrac{x}{2}$ and $\sin x = 2\sin \dfrac{x}{2}\cos \dfrac{x}{2}$. Therefore, we get

$\Rightarrow \tan 2A = \dfrac{{2\left( {\dfrac{{2{{\sin }^2}\dfrac{B}{2}}}{{2\sin \dfrac{B}{2}\cos \dfrac{B}{2}}}} \right)}}{{1 - {{\left( {\dfrac{{2{{\sin }^2}\dfrac{B}{2}}}{{2\sin \dfrac{B}{2}\cos \dfrac{B}{2}}}} \right)}^2}}}$