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

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

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.

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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 $3 c m$ is cut and bent so as to lie along the circumference of a hoop whose radius is $48 c m$ . Find the angle in degrees which is subtended at the centre of hoop.

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

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

Again, it is being placed along a circular hoop of radius $48 c m$ .

The length $(s)$ of the arc = $6 π$

Radius of circle, $r = 48 c m$

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

$\Rightarrow θ = \frac{A r c}{R a d i u s}$

$\Rightarrow θ = \frac{6 π}{48}$

$\Rightarrow θ = \frac{π}{8}$

$\Rightarrow θ = {22.5}^{\circ}$

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

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

$\Rightarrow A = \cos^{2} θ + \sin^{2} θ \sin^{2} θ ⩽ \cos^{2} θ + \sin^{2} θ$

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

$\Rightarrow A = \cos^{2} θ + \sin^{2} θ \sin^{2} θ ⩽ 1. . . . . . . (i)$

Now, $A = \cos^{2} θ + \sin^{4} θ$

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

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

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

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

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

$\Rightarrow A = {(\sin^{2} θ - \frac{1}{2})}^{2} + \frac{4 - 1}{4}$

$\Rightarrow A = {(\sin^{2} θ - \frac{1}{2})}^{2} + \frac{3}{4}$

Therefore, $A = {(\sin^{2} θ - \frac{1}{2})}^{2} + \frac{3}{4} ⩾ \frac{3}{4} . . . . . . (i i)$

Thus, from equation $(i)$ and $(i i)$ , we get

$\Rightarrow \frac{3}{4} ⩽ A ⩽ 1$

Example 3: Find the value of $\sqrt{3} \cos e c 20^{\circ} - \sec 20^{\circ}$

Ans: We have, $\sqrt{3} \cos e c 20^{\circ} - \sec 20^{\circ}$

We can also write it as,

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

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

Multiply and divide numerator by $2$

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

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

$\Rightarrow \frac{2 (\sin 60^{\circ} \cos 20^{\circ} - \cos 60^{\circ} \sin 20^{\circ})}{\sin 20^{\circ} \cos 20^{\circ}}$

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

$\Rightarrow \frac{2 \sin (60^{\circ} - 20^{\circ})}{\sin 20^{\circ} \cos 20^{\circ}}$

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

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

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

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

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

$\Rightarrow 4$

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

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

Take LHS,

$\Rightarrow \sqrt{\frac{1 - \sin θ}{1 + \sin θ}} + \sqrt{\frac{1 + \sin θ}{1 - \sin θ}}$

$\Rightarrow \sqrt{\frac{(1 - \sin θ) (1 - \sin θ)}{(1 + \sin θ) (1 - \sin θ)}} + \sqrt{\frac{(1 + \sin θ) (1 + \sin θ)}{(1 + \sin θ) (1 + \sin θ)}}$

$\Rightarrow \sqrt{\frac{{(1 - \sin θ)}^{2}}{(1^{2} - \sin^{2} θ)}} + \sqrt{\frac{{(1 + \sin θ)}^{2}}{(1^{2} - \sin^{2} θ)}}$

$\Rightarrow \frac{(1 - \sin θ)}{\sqrt{(1^{2} - \sin^{2} θ)}} + \frac{(1 + \sin θ)}{\sqrt{(1^{2} - \sin^{2} θ)}}$

$\Rightarrow \frac{1 - \sin θ + 1 + \sin θ}{\sqrt{(1^{2} - \sin^{2} θ)}}$

$\Rightarrow \frac{2}{\sqrt{(1^{2} - \sin^{2} θ)}}$

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

$\Rightarrow \frac{2}{\sqrt{\cos θ}}$

$\Rightarrow \frac{2}{| \cos θ |}$

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

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

$\Rightarrow \frac{2}{- \cos θ}$

$\Rightarrow - 2 \sec θ$

Hence proved

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

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

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

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

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

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

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

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

$\Rightarrow \frac{\sin 9^{\circ}}{\cos 9^{\circ}} + \frac{\cos 9^{\circ}}{\sin 9^{\circ}} - (\frac{\sin 27^{\circ}}{\cos 27^{\circ}} + \frac{\cos 27^{\circ}}{\sin 27^{\circ}})$

Take LCM

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

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

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

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

Multiply and divide the expression by $2$

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

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

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

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

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

$\Rightarrow \frac{8 (\sqrt{5} + 1) - 8 (\sqrt{5} - 1)}{5 - 1}$

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

$\Rightarrow \frac{16}{4}$

$\Rightarrow 4$

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

Ans: We have, LHS $\frac{\sec 8 θ - 1}{\sec 4 θ - 1}$

Write the above-written expression in terms of $\cos i n e$

$\Rightarrow \frac{\frac{1}{\cos 8 θ} - 1}{\frac{1}{\cos 4 θ} - 1}$

$\Rightarrow \frac{\frac{1 - \cos 8 θ}{\cos 8 θ}}{\frac{1 - \cos 4 θ}{\cos 4 θ}}$

$\Rightarrow \frac{1 - \cos 8 θ}{\cos 8 θ} \times \frac{\cos 4 θ}{1 - \cos 4 θ}$

$\Rightarrow \frac{(1 - \cos 8 θ) \cos 4 θ}{\cos 8 θ (1 - \cos 4 θ)}$

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

$\Rightarrow \frac{2 \sin^{2} 4 θ \cos 4 θ}{\cos 8 θ 2 \sin^{2} 2 θ}$

$\Rightarrow \frac{\sin 4 θ (2 \sin 4 θ \cos 4 θ)}{2 \cos 8 θ \sin^{2} 2 θ}$

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

$\Rightarrow \frac{\sin 4 θ \sin 8 θ}{2 \cos 8 θ \sin^{2} 2 θ}$

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

$\Rightarrow \frac{2 \sin 2 θ \cos 2 θ \sin 8 θ}{2 \cos 8 θ \sin^{2} 2 θ}$

On canceling common terms, we get

$\Rightarrow \frac{\sin 8 θ \cos 2 θ}{\cos 8 θ \sin 2 θ}$

$\Rightarrow \frac{\tan 8 θ}{\tan 2 θ}$

Hence proved

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

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

$\Rightarrow (\sin 5 θ + \sin θ) + \sin 3 θ = 0$

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

$\Rightarrow (2 \sin \frac{5 θ + θ}{2} \cos \frac{5 θ - θ}{2}) + \sin 3 θ = 0$

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

$\Rightarrow \sin 3 θ (2 \cos 2 θ + 1) = 0$

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

$\Rightarrow \sin 3 θ = 0$ or $\cos 2 θ = \frac{- 1}{2}$

Now, $\sin 3 θ = 0$

$\Rightarrow 3 θ = n π, n \in Z$

$\Rightarrow θ = \frac{n π}{3}, n \in Z$

And, $\cos 2 θ = \frac{- 1}{2}$

$\Rightarrow \cos 2 θ = \cos (π - \frac{π}{3})$

$\Rightarrow \cos 2 θ = \cos (\frac{3 π - π}{3})$

$\Rightarrow \cos 2 θ = \cos (\frac{2 π}{3})$

We know that if $\cos θ = \cos α$ , then $θ = 2 m π \pm α$ . Therefore, we get

$\Rightarrow 2 θ = 2 m π \pm \frac{2 π}{3}, m \in Z$

$\Rightarrow θ = m π \pm \frac{π}{3}, m \in Z$

Hence, the general solution of the given equation is $θ = \frac{n π}{3}, n \in Z$ or, $θ = m π \pm \frac{π}{3}, m \in Z$ .

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

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

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

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

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

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

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

$\Rightarrow {(\tan x)}^{2} = {(\frac{1}{\sqrt{3}})}^{2}$

$\Rightarrow {(\tan x)}^{2} = {(\tan \frac{π}{6})}^{2}$

$\Rightarrow \tan^{2} x = \tan^{2} \frac{π}{6}$

We know that if $\tan^{2} θ = \tan^{2} α$ , then $θ = n π \pm α$ , $n \in Z$ .

$\Rightarrow x = n π \pm \frac{π}{6}$

Therefore, possible solutions are $\frac{π}{6}$ , $\frac{5 π}{6}$ , $\frac{7 π}{6}$ , $\frac{11 π}{6}$ where $0 ⩽ x ⩽ 2 π$ .

Long Answer Type

Example 9: Find the value of $(1 + \cos \frac{π}{8}) (1 + \cos \frac{3 π}{8}) (1 + \cos \frac{5 π}{8}) (1 + \cos \frac{7 π}{8})$

Ans: We have, $(1 + \cos \frac{π}{8}) (1 + \cos \frac{3 π}{8}) (1 + \cos \frac{5 π}{8}) (1 + \cos \frac{7 π}{8})$

$\Rightarrow (1 + \cos \frac{π}{8}) (1 + \cos \frac{3 π}{8}) (1 + \cos (π - \frac{3 π}{8})) (1 + \cos (π - \frac{π}{8}))$

We know that $\cos (π - θ) = - \cos θ$ . Therefore, we get

$\Rightarrow (1 + \cos \frac{π}{8}) (1 + \cos \frac{3 π}{8}) (1 - \cos \frac{3 π}{8}) (1 - \cos \frac{π}{8})$

$\Rightarrow (1 - \cos^{2} \frac{π}{8}) (1 - \cos^{2} \frac{3 π}{8})$

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

$\Rightarrow \sin^{2} \frac{π}{8} \sin^{2} \frac{3 π}{8}$

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

$\Rightarrow \frac{1}{4} (2 \sin^{2} \frac{π}{8}) (2 \sin^{2} \frac{3 π}{8})$

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

$\Rightarrow \frac{1}{4} (1 - \cos \frac{π}{4}) (1 - \cos \frac{3 π}{4})$

$\Rightarrow \frac{1}{4} (1 - \frac{1}{\sqrt{2}}) (1 + \frac{1}{\sqrt{2}})$

$\Rightarrow \frac{1}{4} (1^{2} - {(\frac{1}{\sqrt{2}})}^{2})$

$\Rightarrow \frac{1}{4} (1 - \frac{1}{2})$

$\Rightarrow \frac{1}{4} (\frac{1}{2})$

$\Rightarrow \frac{1}{8}$

Example 10: If $x \cos θ = y \cos (θ + \frac{2 π}{3}) = z \cos (θ + \frac{4 π}{3})$ , then find the value of $x y + y z + z x$

Ans: Note that $x y + y z + z x = x y z (\frac{1}{x} + \frac{1}{y} + \frac{1}{z})$

Given that, $x \cos θ = y \cos (θ + \frac{2 π}{3}) = z \cos (θ + \frac{4 π}{3})$

Let $x \cos θ = y \cos (θ + \frac{2 π}{3}) = z \cos (θ + \frac{4 π}{3}) = k$

Then, $x = \frac{k}{\cos θ}$ , $y = \frac{k}{\cos (θ + \frac{2 π}{3})}$ and $z = \frac{k}{\cos (θ + \frac{4 π}{3})}$

Now, we have

$\Rightarrow \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = \frac{1}{\frac{k}{\cos θ}} + \frac{1}{\frac{k}{\cos (θ + \frac{2 π}{3})}} + \frac{1}{\frac{k}{\cos (θ + \frac{4 π}{3})}}$

$\Rightarrow \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = \frac{\cos θ}{k} + \frac{\cos (θ + \frac{2 π}{3})}{k} + \frac{\cos (θ + \frac{4 π}{3})}{k}$

$\Rightarrow \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = \frac{1}{k} [\cos θ + \cos (θ + \frac{2 π}{3}) + \cos (θ + \frac{4 π}{3})]$

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

$\Rightarrow \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = \frac{1}{k} [\cos θ + \cos θ \cos \frac{2 π}{3} - \sin θ \sin \frac{2 π}{3} + \cos θ \cos \frac{4 π}{3} - \sin θ \sin \frac{4 π}{3}]$

$\Rightarrow \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = \frac{1}{k} [\cos θ + \cos θ (\frac{- 1}{2}) - \sin θ (\frac{\sqrt{3}}{2}) + \cos θ (\frac{- 1}{2}) - \sin θ (\frac{- \sqrt{3}}{2})]$

$\Rightarrow \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = \frac{1}{k} [\cos θ - \frac{1}{2} \cos θ - \frac{\sqrt{3}}{2} \sin θ - \frac{1}{2} \cos θ + \frac{\sqrt{3}}{2} \sin θ]$

$\Rightarrow \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = \frac{1}{k} [\cos θ - \cos θ - \frac{\sqrt{3}}{2} \sin θ + \frac{\sqrt{3}}{2} \sin θ]$

$\Rightarrow \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = \frac{1}{k} (0)$

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

$∴ x y + y z + z x = x y z (0)$

$\Rightarrow x y + y z + z x = 0$

Example 11: If $α$ and $β$ are the solutions of the equation $a \tan θ + b \sec θ = c$ , then show that $\tan (α + β) = \frac{2 a c}{a^{2} - c^{2}}$ .

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

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

$\Rightarrow a \frac{\sin θ}{\cos θ} + b \frac{1}{\cos θ} = c$

$\Rightarrow \frac{a \sin θ + b}{\cos θ} = c$

$\Rightarrow a \sin θ + b = c \cos θ$

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

$\Rightarrow \frac{a (2 \tan \frac{θ}{2})}{1 + \tan^{2} \frac{θ}{2}} + b = \frac{c (1 - \tan^{2} \frac{θ}{2})}{1 + \tan^{2} \frac{θ}{2}}$

$\Rightarrow \frac{a (2 \tan \frac{θ}{2}) + b (1 + \tan^{2} \frac{θ}{2})}{1 + \tan^{2} \frac{θ}{2}} = \frac{c (1 - \tan^{2} \frac{θ}{2})}{1 + \tan^{2} \frac{θ}{2}}$

$\Rightarrow 2 a \tan \frac{θ}{2} + b (1 + \tan^{2} \frac{θ}{2}) = c (1 - \tan^{2} \frac{θ}{2})$

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

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

$\Rightarrow (b + c) \tan^{2} \frac{θ}{2} + 2 a \tan \frac{θ}{2} + b - c = 0$

Above written equation is quadratic in $\tan \frac{θ}{2}$ and hence $\tan \frac{α}{2}$ and $\tan \frac{β}{2}$ are the roots of this equation.

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

$\Rightarrow \tan \frac{α}{2} + \tan \frac{β}{2} = \frac{- 2 a}{b + c}$ and $\tan \frac{α}{2} \tan \frac{β}{2} = \frac{b - c}{b + c}$

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

$\Rightarrow \tan (\frac{α}{2} + \frac{β}{2}) = \frac{\tan \frac{α}{2} + \tan \frac{β}{2}}{1 - \tan \frac{α}{2} \tan \frac{β}{2}}$

On substituting the values, we get

$\Rightarrow \tan (\frac{α}{2} + \frac{β}{2}) = \frac{\frac{- 2 a}{b + c}}{1 - (\frac{b - c}{b + c})}$

$\Rightarrow \tan (\frac{α}{2} + \frac{β}{2}) = \frac{\frac{- 2 a}{b + c}}{\frac{b + c - b + c}{b + c}}$

$\Rightarrow \tan (\frac{α}{2} + \frac{β}{2}) = \frac{- 2 a}{2 c}$

$\Rightarrow \tan (\frac{α + β}{2}) = \frac{- a}{c} . . . . . . (i)$

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

$\Rightarrow \tan 2 (\frac{α + β}{2}) = \frac{2 \tan (\frac{α + β}{2})}{1 - \tan^{2} (\frac{α + β}{2})}$

On substituting the values, we get

$\Rightarrow \tan (α + β) = \frac{2 (\frac{- a}{c})}{1 - {(\frac{- a}{c})}^{2}}$

$\Rightarrow \tan (α + β) = \frac{\frac{- 2 a}{c}}{\frac{c^{2} - a^{2}}{c^{2}}}$

$\Rightarrow \tan (α + β) = \frac{- 2 a}{c} \times \frac{c^{2}}{c^{2} - a^{2}}$

$\Rightarrow \tan (α + β) = \frac{- 2 a c}{c^{2} - a^{2}}$

$\Rightarrow \tan (α + β) = \frac{2 a c}{a^{2} - c^{2}}$

Hence proved

Example 12: Show that $2 \sin^{2} β + 4 \cos (α + β) \sin α \sin β + \cos 2 (α + β) = \cos 2 α$

Ans: We have LHS, $2 \sin^{2} β + 4 \cos (α + β) \sin α \sin β + \cos 2 (α + β)$

$2 \sin^{2} β + 4 \cos (α + β) \sin α \sin β + \cos (2 α + 2 β)$

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

$\Rightarrow 2 \sin^{2} β + 4 (\cos α \cos β - \sin α \sin β) \sin α \sin β + (\cos 2 α \cos 2 β - \sin 2 α \sin 2 β)$

$\Rightarrow 2 \sin^{2} β + 4 \sin α \cos α \sin β \cos β - 4 \sin^{2} α \sin^{2} β + \cos 2 α \cos 2 β - \sin 2 α \sin 2 β$

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

$\Rightarrow 2 \sin^{2} β + \sin 2 α \sin 2 β - 4 \sin^{2} α \sin^{2} β + \cos 2 α \cos 2 β - \sin 2 α \sin 2 β$

$\Rightarrow 2 \sin^{2} β - 4 \sin^{2} α \sin^{2} β + \cos 2 α \cos 2 β$

We can also the above written expression as,

$\Rightarrow 2 \sin^{2} β - (2 \sin^{2} α) (2 \sin^{2} β) + \cos 2 α \cos 2 β$

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

$\Rightarrow (1 - \cos 2 β) - (1 - \cos 2 α) (1 - \cos 2 β) + \cos 2 α \cos 2 β$

$\Rightarrow 1 - \cos 2 β - (1 - \cos 2 β - \cos 2 α + \cos 2 β \cos 2 α) + \cos 2 α \cos 2 β$

$\Rightarrow 1 - \cos 2 β - 1 + \cos 2 β + \cos 2 α - \cos 2 β \cos 2 α + \cos 2 α \cos 2 β$

$\Rightarrow \cos 2 α$

Hence proved

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

Ans: Let $θ = α + β$ . Then $\tan α = k \tan β$ .

$\Rightarrow \frac{\tan α}{\tan β} = \frac{k}{1}$

Now, we will apply componendo and dividend

$\Rightarrow \frac{\tan α + \tan β}{\tan α - \tan β} = \frac{k + 1}{k - 1}$

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

$\Rightarrow \frac{\frac{\sin α}{\cos α} + \frac{\sin β}{\cos β}}{\frac{\sin α}{\cos α} - \frac{\sin β}{\cos β}} = \frac{k + 1}{k - 1}$

$\Rightarrow \frac{\frac{\sin α \cos β + \cos α \sin β}{\cos α \cos β}}{\frac{\sin α \cos β - \cos α \sin β}{\cos α \cos β}} = \frac{k + 1}{k - 1}$

$\Rightarrow \frac{\sin α \cos β + \cos α \sin β}{\sin α \cos β - \cos α \sin β} = \frac{k + 1}{k - 1}$

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

$\Rightarrow \frac{\sin (α + β)}{\sin (α - β)} = \frac{k + 1}{k - 1}$

Given that, $α - β = ϕ$ and $α + β = θ$ . Therefore, we get

$\Rightarrow \frac{\sin θ}{\sin ϕ} = \frac{k + 1}{k - 1}$

$\Rightarrow \sin θ = \frac{k + 1}{k - 1} \sin ϕ$

Hence proved

Example 14: Solve $\sqrt{3} \cos θ + \sin θ = \sqrt{2}$ .

Ans: We have equation, $\sqrt{3} \cos θ + \sin θ = \sqrt{2}$

Divide the equation by $2$

$\Rightarrow \frac{\sqrt{3}}{2} \cos θ + \frac{1}{2} \sin θ = \frac{\sqrt{2}}{2}$

$\Rightarrow \frac{\sqrt{3}}{2} \cos θ + \frac{1}{2} \sin θ = \frac{1}{\sqrt{2}}$

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

$\Rightarrow \cos \frac{π}{6} \cos θ + \sin \frac{π}{6} \sin θ = \cos \frac{π}{4}$

$\Leftrightarrow \cos θ \cos \frac{π}{6} + \sin θ \sin \frac{π}{6} = \cos \frac{π}{4}$

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

$\Rightarrow \cos (θ - \frac{π}{6}) = \cos \frac{π}{4}$

We know that when $\cos θ = \cos α$ , then $θ = 2 n π \pm α$ , where $n \in Z$ .

$\Rightarrow θ - \frac{π}{6} = 2 n π \pm \frac{π}{4}$

$\Rightarrow θ = 2 n π \pm \frac{π}{4} + \frac{π}{6}$

Hence, the solutions are $θ = 2 n π + \frac{π}{4} + \frac{π}{6}$ and $θ = 2 n π - \frac{π}{4} + \frac{π}{6}$

i.e., $θ = 2 n π + \frac{5 π}{12}$ and $θ = 2 n π - \frac{π}{12}$

Objective Type Questions

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

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

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

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

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

d) None of these

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

Given that, $\tan θ = \frac{- 4}{3} = \frac{P}{B}$ .

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

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

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

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

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

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

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

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

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

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

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

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

$\Rightarrow \sin θ + \cos θ = \frac{b}{a} . . . (i)$ and $\sin θ \cos θ = \frac{c}{a} . . . . . (i i)$

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

$\Rightarrow \sin^{2} θ + \cos^{2} θ + 2 \sin θ \cos θ = \frac{b^{2}}{a^{2}}$

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

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

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

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

On cross multiplication, we get

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

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

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

a) $1$

b) $2$

c) $\sqrt{2}$

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

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

We have, $\sin x \cos x$

Multiply and divide the expression by $2$

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

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

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

We know that,

$\Rightarrow - 1 ⩽ \sin 2 x ⩽ 1$

Divide the expression by $2$

$\Rightarrow - \frac{1}{2} ⩽ \frac{\sin 2 x}{2} ⩽ \frac{1}{2}$

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

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

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

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

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

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

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

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

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

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

Multiply and divide the expression by $2$ .

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

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

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

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

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

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

Multiply and divide the expression by $2$

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

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

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

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

We know that $\sin (- θ) = - \sin θ$ . Therefore, we get

$\Rightarrow \frac{\sqrt{3}}{8} (\frac{\sqrt{3}}{2})$

$\Rightarrow \frac{3}{16}$

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

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

b) $0$

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

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

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

We have, $\cos \frac{π}{5} \cos \frac{2 π}{5} \cos \frac{4 π}{5} \cos \frac{8 π}{5}$

Multiply the above-written expression by $2 \sin \frac{π}{5}$ .

$\Rightarrow \frac{1}{2 \sin \frac{π}{5}} 2 \sin \frac{π}{5} \cos \frac{π}{5} \cos \frac{2 π}{5} \cos \frac{4 π}{5} \cos \frac{8 π}{5}$

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

$\Rightarrow \frac{1}{2 \sin \frac{π}{5}} \sin \frac{2 π}{5} \cos \frac{2 π}{5} \cos \frac{4 π}{5} \cos \frac{8 π}{5}$

Multiply and divide the expression by $2$

$\Rightarrow \frac{1}{2 \times 2 \sin \frac{π}{5}} 2 \sin \frac{2 π}{5} \cos \frac{2 π}{5} \cos \frac{4 π}{5} \cos \frac{8 π}{5}$

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

$\Rightarrow \frac{1}{4 \sin \frac{π}{5}} \sin \frac{4 π}{5} \cos \frac{4 π}{5} \cos \frac{8 π}{5}$

Multiply and divide the expression by $2$

$\Rightarrow \frac{1}{2 \times 4 \sin \frac{π}{5}} 2 \sin \frac{4 π}{5} \cos \frac{4 π}{5} \cos \frac{8 π}{5}$

$\Rightarrow \frac{1}{8 \sin \frac{π}{5}} \sin \frac{8 π}{5} \cos \frac{8 π}{5}$

Multiply and divide the expression by $2$

$\Rightarrow \frac{1}{2 \times 8 \sin \frac{π}{5}} 2 \sin \frac{8 π}{5} \cos \frac{8 π}{5}$

$\Rightarrow \frac{1}{16 \sin \frac{π}{5}} \sin \frac{16 π}{5}$

$\Rightarrow \frac{\sin (3 π + \frac{π}{5})}{16 \sin \frac{π}{5}}$

$\Rightarrow \frac{- \sin \frac{π}{5}}{16 \sin \frac{π}{5}}$

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

Fill in the blank:

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

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

$\Rightarrow \frac{\tan (θ + 15^{\circ})}{\tan (θ - 15^{\circ})} = \frac{3}{1}$

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

$\Rightarrow \frac{\tan (θ + 15^{\circ}) + \tan (θ - 15^{\circ})}{\tan (θ + 15^{\circ}) - \tan (θ - 15^{\circ})} = \frac{3 + 1}{3 - 1}$

$\Rightarrow \frac{\frac{\sin (θ + 15^{\circ})}{\cos (θ + 15^{\circ})} + \frac{\sin (θ - 15^{\circ})}{\cos (θ - 15^{\circ})}}{\frac{\sin (θ + 15^{\circ})}{\cos (θ + 15^{\circ})} - \frac{\sin (θ - 15^{\circ})}{\cos (θ - 15^{\circ})}} = \frac{4}{2}$

$\Rightarrow \frac{\frac{\sin (θ + 15^{\circ}) \cos (θ - 15^{\circ}) + \sin (θ - 15^{\circ}) \cos (θ + 15^{\circ})}{\cos (θ + 15^{\circ}) \cos (θ - 15^{\circ})}}{\frac{\sin (θ + 15^{\circ}) \cos (θ - 15^{\circ}) - \sin (θ - 15^{\circ}) \cos (θ + 15^{\circ})}{\cos (θ + 15^{\circ}) \cos (θ - 15^{\circ})}} = 2$

$\Rightarrow \frac{\sin (θ + 15^{\circ}) \cos (θ - 15^{\circ}) + \sin (θ - 15^{\circ}) \cos (θ + 15^{\circ})}{\sin (θ + 15^{\circ}) \cos (θ - 15^{\circ}) - \sin (θ - 15^{\circ}) \cos (θ + 15^{\circ})} = 2$

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

$\Rightarrow \frac{\sin (θ + 15^{\circ} + θ - 15^{\circ})}{\sin (θ + 15^{\circ} - (θ - 15^{\circ}))} = 2$

$\Rightarrow \frac{\sin 2 θ}{\sin (θ + 15^{\circ} - θ + 15^{\circ})} = 2$

$\Rightarrow \frac{\sin 2 θ}{\sin 30^{\circ}} = 2$

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

$\Rightarrow \frac{\sin 2 θ}{\frac{1}{2}} = 2$

$\Rightarrow \frac{2 \sin 2 θ}{1} = 2$

$\Rightarrow \sin 2 θ = 1$

Given that, $0^{\circ} < θ < 90^{\circ}$ i.e., $0 < θ < \frac{π}{2}$

$∴ 0 < 2 θ < π$

$\Rightarrow \sin 2 θ = \sin \frac{π}{2}$

$\Rightarrow 2 θ = \frac{π}{2}$

$\Rightarrow θ = \frac{π}{4}$

$\Rightarrow θ = 45^{\circ}$

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

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

Ans: The given statement is true.

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

$\Rightarrow \frac{2^{\sin θ} + 2^{\cos θ}}{2} ⩾ \sqrt{2^{\sin θ} \times 2^{\cos θ}}$

$\Rightarrow 2^{\sin θ} + 2^{\cos θ} ⩾ 2 \sqrt{2^{\sin θ + \cos θ}}$

$\Rightarrow 2^{\sin θ} + 2^{\cos θ} ⩾ {2.2}^{\frac{\sin θ + \cos θ}{2}} . . . . . . (i)$

We have, $\sin θ + \cos θ$ .

Divide and multiply the expression by $\sqrt{2}$

$\Rightarrow \sqrt{2} (\sin θ \times \frac{1}{\sqrt{2}} + \cos θ \times \frac{1}{\sqrt{2}})$

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

$\Rightarrow \sqrt{2} (\sin θ \cos \frac{π}{4} + \cos θ \sin \frac{π}{4})$

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

$\Rightarrow \sqrt{2} \sin (θ + \frac{π}{4})$

Since, $- 1 ⩽ \sin (θ + \frac{π}{4}) ⩽ 1$

$\Rightarrow - \sqrt{2} ⩽ \sqrt{2} \sin (θ + \frac{π}{4}) ⩽ \sqrt{2}$

As we solved above, $\sin θ + \cos θ = \sqrt{2} \sin (θ + \frac{π}{4})$ . Therefore, we get

$\Rightarrow - \sqrt{2} ⩽ \sin θ + \cos θ ⩽ \sqrt{2}$

$\Rightarrow - \frac{\sqrt{2}}{2} ⩽ \frac{\sin θ + \cos θ}{2} ⩽ \frac{\sqrt{2}}{2}$

$\Rightarrow - \frac{1}{\sqrt{2}} ⩽ \frac{\sin θ + \cos θ}{2} ⩽ \frac{1}{\sqrt{2}}$

We have, $2^{\sin θ} + 2^{\cos θ} ⩾ {2.2}^{\frac{\sin θ + \cos θ}{2}} . . . . . . (i)$

$\Rightarrow 2^{\sin θ} + 2^{\cos θ} ⩾ {2.2}^{\frac{- 1}{\sqrt{2}}}$

$\Rightarrow 2^{\sin θ} + 2^{\cos θ} ⩾ 2^{1 + (\frac{- 1}{\sqrt{2}})}$

$\Rightarrow 2^{\sin θ} + 2^{\cos θ} ⩾ 2^{1 - \frac{1}{\sqrt{2}}}$

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

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

c) $\frac{1 + \cos x}{\sin x}$ (iii) $| \cos x + \sin x |$

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

Ans:

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

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

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

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

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

Hence, $(a) \leftrightarrow (i v)$ .

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

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

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

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

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

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

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

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

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

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

Hence, $(c) \leftrightarrow (i i)$

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

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

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

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

$\Rightarrow | (\sin x + \cos x) |$

$\Rightarrow | \cos x + \sin x |$

Hence, $(d) \leftrightarrow (i i i)$ .

Exercise 3.3

Short Answer Type

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

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

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

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

$\Rightarrow \frac{\tan A + \sec A - (\sec^{2} A - \tan^{2} A)}{\tan A - \sec A + 1}$

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

$\Rightarrow \frac{\tan A + \sec A - [(\sec A + \tan A) (\sec A - \tan A)]}{\tan A - \sec A + 1}$

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

$\Rightarrow \frac{(\sec A + \tan A) [1 - (\sec A - \tan A)]}{\tan A - \sec A + 1}$

$\Rightarrow \frac{(\sec A + \tan A) [1 - \sec A + \tan A]}{\tan A - \sec A + 1}$

On canceling common term, we get

$\Rightarrow \sec A + \tan A$

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

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

Take LCM

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

Hence proved.

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

Ans: Given, $y = \frac{2 \sin α}{1 + \cos α + \sin α}$

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

$\Rightarrow \frac{2 \sin α}{1 + \cos α + \sin α} \times \frac{1 + \sin α - \cos α}{1 + \sin α - \cos α}$

$\Rightarrow \frac{2 \sin α (1 + \sin α - \cos α)}{(1 + \cos α + \sin α) (1 + \sin α - \cos α)}$

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

$\Rightarrow \frac{2 \sin α (1 + \sin α - \cos α)}{{(1 + \sin α)}^{2} - \cos^{2} α}$

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

$\Rightarrow \frac{2 \sin α (1 + \sin α - \cos α)}{1^{2} + \sin^{2} α + 2 \sin α - \cos^{2} α}$

$\Rightarrow \frac{2 \sin α (1 + \sin α - \cos α)}{(1^{2} - \cos^{2} α) + \sin^{2} α + 2 \sin α}$

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

$\Rightarrow \frac{2 \sin α (1 + \sin α - \cos α)}{\sin^{2} α + \sin^{2} α + 2 \sin α}$

$\Rightarrow \frac{2 \sin α (1 + \sin α - \cos α)}{2 \sin^{2} α + 2 \sin α}$

Take $2 \sin α$ as a common term

$\Rightarrow \frac{2 \sin α (1 + \sin α - \cos α)}{2 \sin α (\sin α + 1)}$

On canceling common term, we get

$\Rightarrow \frac{1 + \sin α - \cos α}{\sin α + 1} = y$

Hence proved.

If $m \sin θ = n \sin (θ + 2 α)$ , then prove that $\tan (θ + α) \cot α = \frac{m + n}{m - n}$ .

Ans: Given, $m \sin θ = n \sin (θ + 2 α)$

We can also write it as,

$\Rightarrow \frac{\sin (θ + 2 α)}{\sin θ} = \frac{m}{n}$

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

$\Rightarrow \frac{\sin (θ + 2 α) + \sin θ}{\sin (θ + 2 α) - \sin θ} = \frac{m + n}{m - n}$

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

$\Rightarrow \frac{2 \sin (\frac{θ + 2 α + θ}{2}) . \cos (\frac{θ + 2 α - θ}{2})}{2 \cos (\frac{θ + 2 α + θ}{2}) . \sin (\frac{θ + 2 α - θ}{2})} = \frac{m + n}{m - n}$

$\Rightarrow \frac{2 \sin (\frac{2 θ + 2 α}{2}) . \cos (\frac{2 α}{2})}{2 \cos (\frac{2 θ + 2 α}{2}) . \sin (\frac{2 α}{2})} = \frac{m + n}{m - n}$

On simplification, we get

$\Rightarrow \frac{2 \sin (θ + α) . \cos (α)}{2 \cos (θ + α) . \sin (α)} = \frac{m + n}{m - n}$

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

$\Rightarrow \tan (θ + α) . \cot α = \frac{m + n}{m - n}$

Hence proved.

If $\cos (α + β) = \frac{4}{5}$ and $\sin (α - β) = \frac{5}{13}$ , where $α$ lie between $0$ and $\frac{π}{4}$ , find the value of $\tan 2 α$ .

Ans: Given, $\cos (α + β) = \frac{4}{5}$ and $\sin (α - β) = \frac{5}{13}$

At first we will find the value $\tan (α + β)$ and $\tan (α - β)$ .

For value of $\tan (α + β)$ , we have

$\Rightarrow \cos (α + β) = \frac{4}{5} = \frac{B}{H}$

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 $\tan x = \frac{P}{B}$ . Therefore, we get

$\Rightarrow \tan (α + β) = \frac{3}{4}$

For value of $\tan (α - β)$ , we have

$\Rightarrow \sin (α - β) = \frac{5}{13} = \frac{P}{H}$

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 $\tan x = \frac{P}{B}$ . Therefore, we get

$\Rightarrow \tan (α - β) = \frac{5}{12}$

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

$\Rightarrow \tan 2 α = [α + β + α - β]$

$\Rightarrow \tan 2 α = [(α + β) + (α - β)]$

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

$\Rightarrow \tan 2 α = \frac{\tan (α + β) + \tan (α - β)}{1 - \tan (α + β) \tan (α - β)}$

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

$\Rightarrow \tan 2 α = \frac{\frac{9 + 5}{12}}{\frac{48 - 15}{48}}$

$\Rightarrow \tan 2 α = \frac{14}{12} \times \frac{48}{33}$

$\Rightarrow \tan 2 α = \frac{56}{33}$

This is our required answer.

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

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

Let us first take LCM of the above written expression

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

On multiplication of terms, we get

$\Rightarrow \frac{{(\sqrt{a + b})}^{2} + {(\sqrt{a - b})}^{2}}{\sqrt{(a - b) (a + b)}}$

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

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

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

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

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

We can also write it as,

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

On taking LCM, we get

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

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

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

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

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

Prove that $\cos θ \cos \frac{θ}{2} - \cos 3 θ \cos \frac{9 θ}{2} = \sin 4 θ \sin (\frac{7 θ}{2})$ .

Ans: We need to prove that $\cos θ \cos \frac{θ}{2} - \cos 3 θ \cos \frac{9 θ}{2} = \sin 4 θ \sin (\frac{7 θ}{2})$

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

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

$\Rightarrow \frac{1}{2} [2 \cos θ \cos \frac{θ}{2}] - \frac{1}{2} [2 \cos 3 θ \cos \frac{9 θ}{2}]$

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

$\Rightarrow \frac{1}{2} [\cos (θ + \frac{θ}{2}) + \cos (θ - \frac{θ}{2})] - \frac{1}{2} [\cos (3 θ + \frac{9 θ}{2}) + \cos (3 θ - \frac{9 θ}{2})]$

On simplification, we get

$\Rightarrow \frac{1}{2} [\cos \frac{3 θ}{2} + \cos \frac{θ}{2}] - \frac{1}{2} [\cos \frac{15 θ}{2} + \cos (- \frac{3 θ}{2})]$

$\Rightarrow \frac{1}{2} [\cos \frac{3 θ}{2} + \cos \frac{θ}{2} - \cos \frac{15 θ}{2} - \cos (- \frac{3 θ}{2})]$

As we know $\cos (- θ) = \cos θ$ . Therefore, we get

$\Rightarrow \frac{1}{2} [\cos \frac{3 θ}{2} + \cos \frac{θ}{2} - \cos \frac{15 θ}{2} - \cos \frac{3 θ}{2}]$

$\Rightarrow \frac{1}{2} [\cos \frac{θ}{2} - \cos \frac{15 θ}{2}]$

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

$\Rightarrow \frac{1}{2} [- 2 \sin (\frac{\frac{θ}{2} + \frac{15 θ}{2}}{2}) \sin (\frac{\frac{θ}{2} - \frac{15 θ}{2}}{2})]$

$\Rightarrow - \sin (\frac{\frac{16 θ}{2}}{2}) \sin (\frac{\frac{- 14 θ}{2}}{2})$

$\Rightarrow - \sin (4 θ) \sin (\frac{- 7 θ}{2})$

As we know $\sin (- θ) = - \sin θ$ . Therefore, we get

$\Rightarrow \sin (4 θ) \sin (\frac{7 θ}{2})$

Hence proved.

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

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

$\Rightarrow a \cos θ + b \sin θ = m$

On squaring both the side, we get

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

$\Rightarrow a^{2} \cos^{2} θ + b^{2} \sin^{2} θ + 2 a b \sin θ \cos θ = m^{2} . . . . . (i)$

$\Rightarrow a \sin θ - b \cos θ = n$

On squaring both the side, we get

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

$\Rightarrow a^{2} \sin^{2} θ + b^{2} \cos^{2} θ - 2 a b \sin θ \cos θ = n^{2} . . . . . (i i)$

Add equation $(i)$ and $(i i)$ .

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

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

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

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

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

Hence proved.

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

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

Let $22^{\circ} 33^{'} = \frac{θ}{2}$

$\Rightarrow θ = 45^{\circ}$

$\Rightarrow \tan 20^{\circ} 30^{'} = \tan \frac{θ}{2}$

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

$\Rightarrow \tan 20^{\circ} 30^{'} = \frac{\sin \frac{θ}{2}}{\cos \frac{θ}{2}}$

Multiply numerator and denominator by $2 \cos \frac{θ}{2}$

$\Rightarrow \tan 20^{\circ} 30^{'} = \frac{2 \sin \frac{θ}{2} \cos \frac{θ}{2}}{2 \cos \frac{θ}{2} \cos \frac{θ}{2}}$

$\Rightarrow \tan 20^{\circ} 30^{'} = \frac{2 \sin \frac{θ}{2} \cos \frac{θ}{2}}{2 \cos^{2} \frac{θ}{2}}$

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

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

Put $θ = 45^{\circ}$

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

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

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

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

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

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

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

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

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

This is our required answer.

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

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

We can write $\sin 4 A = \sin (A + 3 A)$

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

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

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

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

On multiplication of terms, we get

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

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

Hence proved.

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

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

$\Rightarrow \tan θ + \sin θ = m$

On squaring both the sides, we get

$\Rightarrow {(\tan θ + \sin θ)}^{2} = m^{2}$

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

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

$\Rightarrow \tan θ - \sin θ = n$

On squaring both the sides, we get

$\Rightarrow {(\tan θ - \sin θ)}^{2} = n^{2}$

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

$\Rightarrow \tan^{2} θ + \sin^{2} θ - 2 \tan θ \sin θ = n^{2} . . . . (i i)$

Now, we will subtract equation $(i i)$ from $(i)$ .

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

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

On subtraction of terms, we get

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

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

Hence proved.

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

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

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

$\Rightarrow \tan 2 A = \tan (A + B + A - B)$

This can also be written as,

$\Rightarrow \tan 2 A = \tan [(A + B) + (A - B)]$

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

$\Rightarrow \tan 2 A = \frac{\tan (A + B) + \tan (A - B)}{1 - \tan (A + B) \tan (A - B)}$

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

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

Hence proved.

If $\cos α + \cos β = 0 = \sin α + \sin β$ , then prove that $\cos 2 α + \cos 2 β = - 2 \cos (α + β)$ .

Ans: Given, $\cos α + \cos β = 0$ and $\sin α + \sin β = 0$ .

$\Rightarrow {(\cos α + \cos β)}^{2} = 0$

On squaring both the sides, we get

$\Rightarrow \cos^{2} α + \cos^{2} β + 2 \cos α \cos β = 0. . . . . . . . (1)$

$\Rightarrow \sin α + \sin β = 0$

On squaring both the sides, we get

$\Rightarrow {(\sin α + \sin β)}^{2} = 0$

$\Rightarrow \sin^{2} α + \sin^{2} β + 2 \sin α \sin β = 0. . . . . . . (2)$

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

$\Rightarrow \cos^{2} α + \cos^{2} β + 2 \cos α \cos β - \sin^{2} α - \sin^{2} β - 2 \sin α \sin β = 0$

$\Rightarrow \cos^{2} α - \sin^{2} α + \cos^{2} β - \sin^{2} β + 2 \cos α \cos β - 2 \sin α \sin β = 0$

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

$\Rightarrow \cos 2 α + \cos 2 β + 2 (\cos α \cos β - \sin α \sin β) = 0$

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

$\Rightarrow \cos 2 α + \cos 2 β + 2 \cos (α + β) = 0$

$\Rightarrow \cos 2 α + \cos 2 β = - 2 \cos (α + β)$

Hence proved.

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

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

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

$\Rightarrow \frac{\sin (x + y) + \sin (x - y)}{\sin (x + y) - \sin (x - y)} = \frac{a + b + a + b}{a + b - (a - b)}$

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

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

On simplification, we get

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

On canceling common terms, we get

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

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

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

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

Hence proved.

If $\tan θ = \frac{\sin α - \cos α}{\sin α + \cos α}$ , then show that $\sin α + \cos α = \sqrt{2} \cos θ$ .

Ans: Given, $\tan θ = \frac{\sin α - \cos α}{\sin α + \cos α}$

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

$\Rightarrow \tan θ = \frac{\frac{\sin α - \cos α}{\cos α}}{\frac{\sin α + \cos α}{\cos α}}$

$\Rightarrow \tan θ = \frac{\frac{\sin α}{\cos α} - \frac{\cos α}{\cos α}}{\frac{\sin α}{\cos α} + \frac{\cos α}{\cos α}}$

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

$\Rightarrow \tan θ = \frac{\tan α - 1}{\tan α + 1}$

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

$\Rightarrow \tan θ = \frac{\tan α - \tan \frac{π}{4}}{\tan α + \tan \frac{π}{4}}$

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

$\Rightarrow \tan θ = \tan (α - \frac{π}{4})$

$∴ θ = α - \frac{π}{4}$

$\Rightarrow \cos θ = \cos (α - \frac{π}{4})$

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

$\Rightarrow \cos θ = \cos α . \cos \frac{π}{4} + \sin α . \sin \frac{π}{4}$

Substitute values of $\cos \frac{π}{4} = \frac{1}{\sqrt{2}}$ and $\sin \frac{π}{4} = \frac{1}{\sqrt{2}}$

$\Rightarrow \cos θ = \cos α . \frac{1}{\sqrt{2}} + \sin α . \frac{1}{\sqrt{2}}$

Multiply both sides by $\sqrt{2}$ .

$\Rightarrow \sqrt{2} \cos θ = \cos α + \sin α$

$\Rightarrow \sin α + \cos α = \sqrt{2} \cos θ$

Hence proved.

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

Ans: We have, $\sin θ + \cos θ = 1$

Divide both sides by $\sqrt{2}$ .

$\Rightarrow \frac{1}{\sqrt{2}} \sin θ + \frac{1}{\sqrt{2}} \cos θ = \frac{1}{\sqrt{2}}$

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

$\Rightarrow \sin \frac{π}{4} \sin θ + \cos \frac{π}{4} \cos θ = \frac{1}{\sqrt{2}}$

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

$\Rightarrow \cos (θ - \frac{π}{4}) = \cos \frac{π}{4}$

We know that if $\cos θ = \cos α$ , then $θ = 2 n π \pm α$ . Therefore, we get

$\Rightarrow θ - \frac{π}{4} = 2 n π \pm \frac{π}{4}, n \in Z$

Transport $- \frac{π}{4}$ to RHS

$\Rightarrow θ = 2 n π \pm \frac{π}{4} - \frac{π}{4}$

$\Rightarrow θ = 2 n π + \frac{π}{4} + \frac{π}{4}$ or $\Rightarrow θ = 2 n π - \frac{π}{4} + \frac{π}{4}$

$\Rightarrow θ = 2 n π + \frac{π}{2}$ or $\Rightarrow θ = 2 n π, n \in Z$

Therefore, the general values of $θ$ are $2 n π + \frac{π}{2}$ and $2 n π$ where $n \in Z$ .

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

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

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

We have, $\tan θ = - 1$

$\Rightarrow \tan θ = \tan (- \frac{π}{4})$

We know that $\tan (2 π - x) = - \tan x$ . So, we can write above written expression as

$\Rightarrow \tan θ = \tan (2 π - \frac{π}{4})$

$\Rightarrow \tan θ = \tan (\frac{7 π}{4})$

$\Rightarrow θ = \frac{7 π}{4}$

We have, $\cos θ = \frac{1}{\sqrt{2}}$

$\Rightarrow \cos θ = \cos \frac{π}{4}$

We know that $\cos (2 π - x) = \cos x$ . So, we can write above written expression as

$\Rightarrow \cos θ = \cos (2 π - \frac{π}{4})$

$\Leftarrow \cos θ = \cos \frac{7 π}{4}$

$\Rightarrow θ = \frac{7 π}{4}$

Therefore, the general solution is $θ = 2 n π + \frac{7 π}{4}, n \in Z$ .

If $\cot θ + \tan θ = 2 \cos e c θ$ , then find the general value of $θ$ .

Ans: Given, $\cot θ + \tan θ = 2 \cos e c θ$

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

$\Rightarrow \frac{\cos θ}{\sin θ} + \frac{\sin θ}{\cos θ} = \frac{2}{\sin θ}$

Take LCM

$\Rightarrow \frac{\cos^{2} θ + \sin^{2} θ}{\sin θ \cos θ} = \frac{2}{\sin θ}$

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

$\Rightarrow \frac{1}{\sin θ \cos θ} = \frac{2}{\sin θ}$

On cross multiplication, we get

$\Rightarrow \frac{1}{\sin θ \cos θ} = \frac{2}{\sin θ}$

$\Rightarrow 2 \sin θ \cos θ = \sin θ$

Transport $\sin θ$ to LHS

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

$\Rightarrow \sin θ (2 \cos θ - 1) = 0$

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

$\Rightarrow \sin θ = 0$ or $\cos θ = \frac{1}{2}$

As we have, $\sin θ = 0$

$\Rightarrow θ = n π, n \in Z$

As we have, $\cos θ = \frac{1}{2}$

$\Rightarrow \cos θ = \cos \frac{π}{3}$

$\Rightarrow θ = 2 n π \pm \frac{π}{3}$

Therefore, the general value of $θ$ is $2 n π \pm \frac{π}{3}$ and $n π$ , $n \in Z$ .

If $2 \sin^{2} θ = 3 \cos θ$ , where $0 ⩽ θ ⩽ 2 π$ , then find the value of $θ$ .

Ans: Given, $2 \sin^{2} θ = 3 \cos θ$

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

$\Rightarrow 2 (1 - \cos^{2} θ) = 3 \cos θ$

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

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

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

$\Rightarrow 2 \cos θ (\cos θ + 2) - 1 (\cos θ + 2) = 0$

$\Rightarrow (\cos θ + 2) (2 \cos θ - 1) = 0$

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

$\Rightarrow \cos θ \neq - 2$ $[- 1 ⩽ \cos θ ⩽ 1]$

Therefore, $2 \cos θ - 1 = 0$

$\Rightarrow \cos θ = \frac{1}{2}$

$\Rightarrow \cos θ = \cos \frac{π}{3}$

$\Rightarrow θ = \frac{π}{3}$ or $2 π - \frac{π}{3}$

$\Rightarrow θ = \frac{π}{3}$ or $\frac{5 π}{3}$ ( As it is mentioned in the question that $0 ⩽ θ ⩽ 2 π$ )

Therefore, the value of $θ$ are $\frac{π}{3}$ and $\frac{5 π}{3}$ .

If $\sec x \cos 5 x + 1 = 0$ , where $0 < x ⩽ \frac{π}{2}$ , then find the value of $x$ .

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

We can also write it as,

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

Take LCM

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

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

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

$\Rightarrow 2 \cos (\frac{5 x + x}{2}) \cos (\frac{5 x - x}{2}) = 0$

$\Rightarrow \cos (\frac{6 x}{2}) \cos (\frac{4 x}{2}) = 0$

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

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

$\Rightarrow 3 x = \frac{π}{2}$ or $2 x = \frac{π}{2}$

$\Rightarrow x = \frac{π}{6}$ or $x = \frac{π}{4}$

Therefore, the value of $x$ are $\frac{π}{6}$ , $\frac{π}{4}$ .

Long Answer Type

If $\sin (θ + α) = a$ and $\sin (θ + β) = b$ , then prove that $\cos 2 (α - β) - 4 a b \cos (α - β) = 1 - 2 a^{2} - 2 b^{2}$ .

Ans: Given that, $\sin (θ + α) = a$ and $\sin (θ + β) = b$ .

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

$\Rightarrow \cos (θ + α) = \sqrt{1 - {(\sin (θ + α))}^{2}}$

$\Rightarrow \cos (θ + α) = \sqrt{1 - a^{2}}$

Similarly,

$\Rightarrow \cos (θ + β) = \sqrt{1 - {(\sin (θ + β))}^{2}}$

$\Rightarrow \cos (θ + β) = \sqrt{1 - b^{2}}$

Now, let us find value of $\cos (α - β)$ .

$\Rightarrow \cos (α - β) = \cos [(θ + α) - (θ + β)]$

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

$\Rightarrow \cos (α - β) = \cos (θ + α) \cos (θ + β) + \sin (θ + α) \sin (θ + β)$

On substituting the values, we get

$\Rightarrow \cos (α - β) = \sqrt{1 - a^{2}} \sqrt{1 - b^{2}} + a b$

$\Rightarrow \cos (α - β) = a b + \sqrt{1 - b^{2} - a^{2} + a^{2} b^{2}}$

Now, we will find the value of $\cos 2 (α - β)$

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

$\Rightarrow \cos 2 (α - β) = 2 \cos^{2} (α - β) - 1$

$\Rightarrow \cos 2 (α - β) = 2 {(a b + \sqrt{1 - b^{2} - a^{2} + a^{2} b^{2}})}^{2} - 1$

$\Rightarrow \cos 2 (α - β) = 2 (a^{2} b^{2} + 1 - b^{2} - a^{2} + a^{2} b^{2} + 2 a b \sqrt{1 - b^{2} - a^{2} + a^{2} b^{2}}) - 1$

We have, L.H.S. = $\cos 2 (α - β) - 4 a b \cos (α - β)$ . On substituting the values, we get

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

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

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

On subtraction, we get

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

Hence proved.

If $\cos (θ + ϕ) = m \cos (θ - ϕ)$ , then prove that $\tan θ = \frac{1 - m}{1 + m} \cot ϕ$ .

Ans: Given, $\cos (θ + ϕ) = m \cos (θ - ϕ)$

We can also write it as,

$\Rightarrow \frac{\cos (θ + ϕ)}{\cos (θ - ϕ)} = \frac{m}{1}$

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

$\Rightarrow \frac{\cos (θ + ϕ) + \cos (θ - ϕ)}{\cos (θ + ϕ) - \cos (θ - ϕ)} = \frac{m + 1}{m - 1}$

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

$\Rightarrow \frac{2 \cos (\frac{θ + ϕ + θ - ϕ}{2}) . \cos (\frac{θ + ϕ - θ + ϕ}{2})}{- 2 \sin (\frac{θ + ϕ + θ - ϕ}{2}) . \sin (\frac{θ + ϕ - θ + ϕ}{2})} = \frac{m + 1}{m - 1}$

$\Rightarrow \frac{2 \cos (\frac{2 θ}{2}) . \cos (\frac{2 ϕ}{2})}{- 2 \sin (\frac{2 θ}{2}) . \sin (\frac{2 ϕ}{2})} = \frac{m + 1}{m - 1}$

On canceling common terms, we get

$\Rightarrow \frac{\cos (θ) . \cos (ϕ)}{- \sin (θ) . \sin (ϕ)} = \frac{m + 1}{m - 1}$

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

$\Rightarrow - \cot θ . \cot ϕ = \frac{m + 1}{m - 1}$

$\Rightarrow - \frac{\cot ϕ}{\tan θ} = \frac{m + 1}{m - 1}$

On cross multiplication, we get

$\Rightarrow \tan θ (1 + m) = \cot ϕ (1 - m)$

$\Rightarrow \tan θ = \frac{(1 - m)}{(1 + m)} \cot ϕ$

Hence proved.

Find the value of the expression $3 [\sin^{4} (\frac{3 π}{2} - α) + \sin^{4} (3 π + α)] - 2 [\sin^{6} (\frac{π}{2} + α) + \sin^{6} (5 π - α)]$

Ans: Given, $3 [\sin^{4} (\frac{3 π}{2} - α) + \sin^{4} (3 π + α)] - 2 [\sin^{6} (\frac{π}{2} + α) + \sin^{6} (5 π - α)]$

We can also write it as,

$\Rightarrow 3 [\sin^{4} (\frac{3 π}{2} - α) + \sin^{4} (2 π + (π + α))] - 2 [\sin^{6} (\frac{π}{2} + α) + \sin^{6} (4 π + (π - α))]$

As we know $\sin (\frac{3 π}{2} - x) = - \cos x$ , $\sin (2 π + x) = \sin x$ , $\sin (\frac{π}{2} + x) = \cos x$ and $\sin (4 π + x) = \sin x$ . Therefore, we get

$\Rightarrow 3 [\cos^{4} α + \sin^{4} (π + α)] - 2 [\cos^{6} α + \sin^{6} (π - α)]$

As we know $\sin (π + α) = - \sin α$ and $\sin (π - α) = \sin α$ . Therefore, we get

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

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

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

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

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

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

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

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

From $(i i)$ , we get

$\Rightarrow 3 (1 - 2 \cos^{2} α \sin^{2} α) - 2 ({(\cos^{2} α + \sin^{2} α)}^{2} - 2 \cos^{2} α \sin^{2} α - \cos^{2} α \sin^{2} α)$

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

$\Rightarrow 3 (1 - 2 \cos^{2} α \sin^{2} α) - 2 (1 - 2 \cos^{2} α \sin^{2} α - \cos^{2} α \sin^{2} α)$

$\Rightarrow 3 (1 - 2 \cos^{2} α \sin^{2} α) - 2 (1 - 3 \cos^{2} α \sin^{2} α)$

On multiplication, we get

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

On subtraction, we get

$\Rightarrow 1$

Hence, value of $3 [\sin^{4} (\frac{3 π}{2} - α) + \sin^{4} (3 π + α)] - 2 [\sin^{6} (\frac{π}{2} + α) + \sin^{6} (5 π - α)]$ is $1$ .

If $a \cos 2 θ + b \sin 2 θ = c$ has $α$ and $β$ as its roots, then prove that $\tan α + \tan β = \frac{2 b}{a + c}$ .

Ans: Given, $a \cos 2 θ + b \sin 2 θ = c$ has $α$ and $β$ as its roots.

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

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

$\Rightarrow a (\frac{1 - \tan^{2} θ}{1 + \tan^{2} θ}) + b (\frac{2 \tan θ}{1 + \tan^{2} θ}) = c$

Take LCM

$\Rightarrow \frac{a (1 - \tan^{2} θ) + b (2 \tan θ)}{1 + \tan^{2} θ} = c$

On cross multiplication, we get

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

On multiplication of terms, we get

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

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

Divide whole equation by $- 1$ .

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

$\Rightarrow (a + c) \tan^{2} θ - 2 b \tan θ + (c - a) = 0. . . . . . . (i i)$

We are given that $α$ and $β$ are the roots of equation $(i)$ . Thus, $\tan α$ and $\tan β$ are the roots of the equation $(i i)$ .

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

$\Rightarrow \tan α + \tan β = \frac{- (- 2 b)}{a + c}$

$\Rightarrow \tan α + \tan β = \frac{2 b}{a + c}$

Hence proved.

If $x = \sec ϕ - \tan ϕ$ and $y = \cos e c ϕ + \cot ϕ$ then show that $x y + x - y + 1 = 0$ .

Ans: Given, $x = \sec ϕ - \tan ϕ$ and $y = \cos e c ϕ + \cot ϕ$

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

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

$\Rightarrow (\sec ϕ - \tan ϕ) (\cos e c ϕ + \cot ϕ) + (\sec ϕ - \tan ϕ) - (\cos e c ϕ + \cot ϕ) + 1$

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

$\Rightarrow (\frac{1 - \sin ϕ}{\cos ϕ}) (\frac{1 + \cos ϕ}{\sin ϕ}) + (\frac{1 - \sin ϕ}{\cos ϕ}) - (\frac{1 + \cos ϕ}{\sin ϕ}) + 1$

$\Rightarrow \frac{(1 - \sin ϕ) (1 + \cos ϕ)}{\cos ϕ \sin ϕ} + (\frac{\sin ϕ (1 - \sin ϕ) - \cos ϕ (1 + \cos ϕ)}{\cos ϕ \sin ϕ}) + 1$

$\Rightarrow \frac{1 - \sin ϕ + \cos ϕ - \sin ϕ \cos ϕ}{\cos ϕ \sin ϕ} + (\frac{\sin ϕ - \sin^{2} ϕ - \cos ϕ - \cos^{2} ϕ}{\cos ϕ \sin ϕ}) + 1$

$\Rightarrow \frac{1 - \sin ϕ + \cos ϕ - \sin ϕ \cos ϕ + \sin ϕ - \cos ϕ - (\sin^{2} ϕ + \cos^{2} ϕ) + \sin ϕ \cos ϕ}{\cos ϕ \sin ϕ}$

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

$\Rightarrow \frac{1 - 1}{\cos ϕ \sin ϕ} = 0$

Hence proved.

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

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

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

$\Rightarrow \sin θ = \sqrt{1 - {(\frac{8}{17})}^{2}}$

$\Rightarrow \sin θ = \sqrt{1 - \frac{64}{289}}$

$\Rightarrow \sin θ = \sqrt{\frac{289 - 64}{289}}$

$\Rightarrow \sin θ = \sqrt{\frac{225}{289}}$

$\Rightarrow \sin θ = \pm \frac{15}{17}$

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

$\Rightarrow \sin θ = \frac{15}{17}$

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

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

$\Rightarrow \cos 30^{\circ} \cos θ - \sin 30^{\circ} \sin θ + \cos 45^{\circ} \cos θ + \sin 45^{\circ} \sin θ + \cos 120^{\circ} \cos θ + \sin 120^{\circ} \sin θ$

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

$\Rightarrow \frac{\sqrt{3}}{2} \times \cos θ - \frac{1}{2} \times \sin θ + \frac{1}{\sqrt{2}} \times \cos θ + \frac{1}{\sqrt{2}} \times \sin θ - \frac{1}{2} \cos θ + \frac{\sqrt{3}}{2} \sin θ$

$\Rightarrow \frac{\sqrt{3}}{2} (\cos θ + \sin θ) - \frac{1}{2} (\cos θ + \sin θ) + \frac{1}{\sqrt{2}} (\cos θ + \sin θ)$

$\Rightarrow (\frac{\sqrt{3}}{2} - \frac{1}{2} + \frac{1}{\sqrt{2}}) (\cos θ + \sin θ)$

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

$\Rightarrow (\frac{\sqrt{3} - 1 + \sqrt{2}}{2}) (\frac{23}{17})$

$\Rightarrow \frac{23}{34} (\sqrt{3} - 1 + \sqrt{2})$

This is our required answer.

Find the value of the expression $\cos^{4} \frac{π}{8} + \cos^{4} \frac{3 π}{8} + \cos^{4} \frac{5 π}{8} + \cos^{4} \frac{7 π}{8}$ .

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

This can also be written as,

$\Rightarrow \cos^{4} \frac{π}{8} + \cos^{4} \frac{3 π}{8} + \cos^{4} (π - \frac{3 π}{8}) + \cos^{4} (π - \frac{π}{8})$

We know that $\cos (π - θ) = - \cos θ$ . Therefore, we get

$\Rightarrow \cos^{4} \frac{π}{8} + \cos^{4} \frac{3 π}{8} + \cos^{4} \frac{3 π}{8} + \cos^{4} \frac{π}{8}$

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

$\Rightarrow 2 \cos^{4} \frac{π}{8} + 2 \cos^{4} \frac{3 π}{8}$

$\Rightarrow 2 [\cos^{4} \frac{π}{8} + \cos^{4} \frac{3 π}{8}]$

$\Rightarrow 2 [\cos^{4} \frac{π}{8} + \cos^{4} (\frac{π}{2} - \frac{π}{8})]$

We know that $\cos (\frac{π}{2} - θ) = \sin θ$ . Therefore, we get

$\Rightarrow 2 [\cos^{4} \frac{π}{8} + \sin^{4} \frac{π}{8}]$

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

$\Rightarrow 2 [{(\cos^{2} \frac{π}{8} + \sin^{2} \frac{π}{8})}^{2} - 2 \sin^{2} \frac{π}{8} \times \cos^{2} \frac{π}{8}]$

$\Rightarrow 2 [1 - 2 \sin^{2} \frac{π}{8} \times \cos^{2} \frac{π}{8}]$

$\Rightarrow 2 - 4 \sin^{2} \frac{π}{8} \times \cos^{2} \frac{π}{8}$

$\Rightarrow 2 - {(2 \sin \frac{π}{8} \times \cos \frac{π}{8})}^{2}$

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

$\Rightarrow 2 - {(\sin \frac{π}{4})}^{2}$

Substitute $\sin \frac{π}{4} = \frac{1}{\sqrt{2}}$ .

$\Rightarrow 2 - {(\frac{1}{\sqrt{2}})}^{2}$

$\Rightarrow 2 - \frac{1}{2} = \frac{4 - 1}{2}$

$\Rightarrow \frac{3}{2}$

Therefore, value of given expression is $\frac{3}{2}$ .

Find the general solution of the equation $5 \cos^{2} θ + 7 \sin^{2} θ - 6 = 0$ .

Ans: Given, $5 \cos^{2} θ + 7 \sin^{2} θ - 6 = 0$

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

$\Rightarrow 5 \cos^{2} θ + 7 (1 - \cos^{2} x) - 6 = 0$

$\Rightarrow 5 \cos^{2} θ + 7 - 7 \cos^{2} x - 6 = 0$

$\Rightarrow 1 - 2 \cos^{2} x = 0$

$\Rightarrow 2 \cos^{2} x = 1$

$\Rightarrow \cos^{2} x = \frac{1}{2}$

$\Rightarrow \cos^{2} x = \cos \frac{π}{4}$

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

$\Rightarrow \frac{1 + \cos 2 θ}{2} = \frac{1 + \cos 2 \times \frac{π}{4}}{2}$

$\Rightarrow 1 + \cos 2 θ = 1 + \cos \frac{π}{2}$

$\Rightarrow \cos 2 θ = \cos \frac{π}{2}$

We know that if $\cos θ = \cos α$ then $θ = 2 n π \pm α$ , $n \in Z$

$\Rightarrow 2 θ = 2 n π \pm \frac{π}{2}$

$\Rightarrow θ = n π \pm \frac{π}{4}$

Therefore, the general solution given equation is $θ = n π \pm \frac{π}{4}$ where $n \in Z$ .

Find the general solution of the equation $\sin x - 3 \sin 2 x + \sin 3 x = \cos x - 3 \cos 2 x + \cos 3 x$ .

Ans: Given equation, $\sin x - 3 \sin 2 x + \sin 3 x = \cos x - 3 \cos 2 x + \cos 3 x$

$\Rightarrow (\sin 3 x + \sin x) - 3 \sin 2 x = (\cos 3 x + \cos x) - 3 \cos 2 x$

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

$\Rightarrow 2 \sin \frac{3 x + x}{2} \cos \frac{3 x - x}{2} - 3 \sin 2 x = 2 \cos \frac{3 x + x}{2} \cos \frac{3 x - x}{2} - 3 \cos 2 x$

$\Rightarrow 2 \sin 2 x . \cos x - 3 \sin 2 x = 2 \cos 2 x . \cos x - 3 \cos 2 x$

$\Rightarrow 2 \cos x (\sin 2 x - \cos 2 x) = 3 (\sin 2 x - \cos 2 x)$

$\Rightarrow 2 \cos x (\sin 2 x - \cos 2 x) - 3 (\sin 2 x - \cos 2 x) = 0$

$\Rightarrow (\sin 2 x - \cos 2 x) (2 \cos x - 3) = 0$

$\Rightarrow \sin 2 x - \cos 2 x = 0$ or $\Rightarrow 2 \cos x - 3 \neq 0$ ( $- 1 ⩽ \cos x ⩽ 1$ )

$\Rightarrow \sin 2 x = \cos 2 x$

$\Rightarrow \frac{\sin 2 x}{\cos 2 x} = 1$

$\Rightarrow \tan 2 x = 1$

We know that if $\tan θ = \tan α$ then $θ = n π + α$ , $n \in Z$

$\Rightarrow \tan 2 x = \tan \frac{π}{4}$

$\Rightarrow 2 x = n π + \frac{π}{4}$

$\Rightarrow x = \frac{n π}{2} + \frac{π}{8}$

Therefore, the general solution given equation is $x = \frac{n π}{2} + \frac{π}{8}$ where $n \in Z$ .

Find the general solution of the equation $(\sqrt{3} - 1) \cos θ + (\sqrt{3} + 1) \sin θ = 2$ .

Ans: Given, $(\sqrt{3} - 1) \cos θ + (\sqrt{3} + 1) \sin θ = 2$

Put $\sqrt{3} - 1 = r \sin α . . . . . . (i)$ and $\sqrt{3} + 1 = r \cos α . . . . . . (i i)$

On squaring and adding both the sides, we get

$\Rightarrow {(r \sin α)}^{2} + {(r \cos α)}^{2} = {(\sqrt{3} - 1)}^{2} + {(\sqrt{3} + 1)}^{2}$

$\Rightarrow r^{2} \sin^{2} α + r^{2} \cos^{2} α = 3 + 1 - 2 \sqrt{3} + 3 + 1 + 2 \sqrt{3}$

$\Rightarrow r^{2} (\sin^{2} α + \cos^{2} α) = 8$

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,

$\Rightarrow r \sin α . \cos θ + r \cos α . \sin θ = 2$

$\Rightarrow r (\sin α . \cos θ + \cos α . \sin θ) = 2$

We know that $\sin (A + B) = \sin A \cos B + \cos A \sin B$ . Therefore, we get

$\Rightarrow 2 \sqrt{2} \sin (α + θ) = 2$

$\Rightarrow \sqrt{2} \sin (α + θ) = 1$

$\Rightarrow \sin (α + θ) = \frac{1}{\sqrt{2}}$

$\Rightarrow \sin (α + θ) = \sin \frac{π}{4}$

We know that if $\sin θ = \sin α$ then $θ = n π + {(- 1)}^{n} α$ , $n \in Z$

$\Rightarrow α + θ = n π + {(- 1)}^{n} . \frac{π}{4} . . . . . . . (i i i)$

We have, $\frac{r \sin α}{r \cos α} = \frac{\sqrt{3} - 1}{\sqrt{3} + 1}$ from equation $(i)$ and $(i i)$ .

This can also be written as, $\tan α = \frac{\sqrt{3} - 1}{1 + \sqrt{3} .1}$

We know that $\tan \frac{π}{3} = \sqrt{3}$ and $\tan \frac{π}{4} = 1$ . Therefore, we get

$\Rightarrow \tan α = \frac{\tan \frac{π}{3} - \tan \frac{π}{4}}{1 + \tan \frac{π}{3} . \tan \frac{π}{4}}$

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

$\Rightarrow \tan α = \tan (\frac{π}{3} - \frac{π}{4})$

$\Rightarrow \tan α = \tan \frac{π}{12}$

$\Rightarrow α = \frac{π}{12}$

Now, we will substitute value of $α$ , in equation $(i i i)$ .

$\Rightarrow \frac{π}{12} + θ = n π + {(- 1)}^{n} . \frac{π}{4}$

$\Rightarrow θ = n π + {(- 1)}^{n} . \frac{π}{4} - \frac{π}{12}$

Therefore, the general solution given equation is $θ = n π + {(- 1)}^{n} . \frac{π}{4} - \frac{π}{12}$ where $n \in Z$ .

OBJECTIVE TYPE QUESTIONS

Choose the correct answer from the given options in the exercises $30$ to $59$ .

If $\sin θ + \cos e c θ = 2$ , then $\sin^{2} θ + \cos e c^{2} θ$ is equal to

$1$
$4$
$2$
None of these

Ans: Given, $\sin θ + \cos e c θ = 2$

On squaring both the sides, we get

$\Rightarrow {(\sin θ + \cos e c θ)}^{2} = 2^{2}$

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

$\Rightarrow \sin^{2} θ + \cos e c^{2} θ + 2 \sin θ \cos e c θ = 4$

$\Rightarrow \sin^{2} θ + \cos e c^{2} θ + 2 \sin θ \frac{1}{\sin θ} = 4$

On canceling common terms, we get

$\Rightarrow \sin^{2} θ + \cos e c^{2} θ + 2 = 4$

Transport $2$ to the RHS

$\Rightarrow \sin^{2} θ + \cos e c^{2} θ = 4 - 2$

$\Rightarrow \sin^{2} θ + \cos e c^{2} θ = 2$

Hence, option (c) is the correct answer.

If $f (x) = \cos^{2} x + \sec^{2} x$ , then

$f (x) < 1$
$f (x) = 1$
$2 < f (x) < 1$
$f (x) ⩾ 2$

Ans: Given, $f (x) = \cos^{2} x + \sec^{2} x$

As we know $A M ⩾ G M$ . Therefore, we get

$\Rightarrow \frac{\cos^{2} x + \sec^{2} x}{2} ⩾ \sqrt{\cos^{2} x \times \sec^{2} x}$

We know that $\cos x = \frac{1}{\sec x}$ . Therefore, we get

$\Rightarrow \frac{\cos^{2} x + \sec^{2} x}{2} ⩾ 1$

On cross multiplication, we get

$\Rightarrow \cos^{2} x + \sec^{2} x ⩾ 2$

$\Rightarrow f (x) ⩾ 2$

Hence, option (d) is the correct answer.

If $\tan θ = \frac{1}{2}$ and $\tan ϕ = \frac{1}{3}$ , then the value of $θ + ϕ$ is

$\frac{π}{6}$
$π$
$0$
$\frac{π}{4}$

Ans: Given, $\tan θ = \frac{1}{2}$ and $\tan ϕ = \frac{1}{3}$

We know that $\tan (θ + ϕ) = \frac{\tan θ + \tan ϕ}{1 - \tan θ \tan ϕ}$ . Therefore, we will substitute the given values in this formula.

$\Rightarrow \tan (θ + ϕ) = \frac{\frac{1}{2} + \frac{1}{3}}{1 - \frac{1}{2} \times \frac{1}{3}}$

On taking LCM, we get

$\Rightarrow \tan (θ + ϕ) = \frac{\frac{3 + 2}{6}}{\frac{6 - 1}{6}}$

$\Rightarrow \tan (θ + ϕ) = \frac{\frac{5}{6}}{\frac{5}{6}} = 1$

As we know $\tan \frac{π}{4} = 1$ . Therefore, we get

$\Rightarrow \tan (θ + ϕ) = \tan \frac{π}{4}$

$\Rightarrow θ + ϕ = \frac{π}{4}$

Hence, option (d) is the correct answer.

Which of the following is not correct?

$\sin θ = - \frac{1}{5}$
$\cos θ = 1$
$\sec θ = \frac{1}{2}$
$\tan θ = 20$

Ans:

As we know $- 1 ⩽ \sin θ ⩽ 1$ . Therefore, $\sin θ = - \frac{1}{5}$ is correct.

We know that $\cos 0^{\circ} = 1$ . Therefore, $\cos θ = 1$ is correct.

We have, $\sec θ = \frac{1}{2}$ . This can also be written as

$\Rightarrow \cos θ = 2$

Which is not correct because $- 1 ⩽ \sin θ ⩽ 1$ .

Hence, option (d) is the correct answer.

The value of $\tan 1^{\circ} \tan 2^{\circ} \tan 3^{\circ} . . . . . \tan 89^{\circ}$ is

$0$
$1$
$\frac{1}{2}$
Not defined

Ans: We need to find the value of $\tan 1^{\circ} \tan 2^{\circ} \tan 3^{\circ} . . . . . \tan 89^{\circ}$

$\Rightarrow \tan 1^{\circ} \tan 2^{\circ} . . . . . \tan 44^{\circ} \tan 45^{\circ} \tan (90^{\circ} - 44^{\circ}) . . . . . . . \tan (90^{\circ} - 2^{\circ}) \tan (90^{\circ} - 1^{\circ})$

We know that $\tan (90^{\circ} - θ) = \cot θ$ . Therefore, we get

$\Rightarrow [\tan 1^{\circ} \tan 2^{\circ} . . . . . \tan 44^{\circ}] \tan 45^{\circ} [\cot 44^{\circ} . . . . . . \cot 2^{\circ} . \cot 1^{\circ}]$

$\Rightarrow [(\tan 1^{\circ} \times \cot 1^{\circ}) (\tan 2^{\circ} \cot 2^{\circ}) . . . . . (\tan 44^{\circ} \cot 44^{\circ})] \times \tan 45^{\circ}$

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.

The value of $\frac{1 - \tan^{2} 15^{\circ}}{1 + \tan^{2} 15^{\circ}}$ is

$1$
$\sqrt{3}$
$\frac{\sqrt{3}}{2}$
$2$

Ans: We have to find the value of $\frac{1 - \tan^{2} 15^{\circ}}{1 + \tan^{2} 15^{\circ}}$

Let $θ = 15^{\circ}$

$\Rightarrow 2 θ = 30^{\circ}$

As we know $\cos 2 θ = \frac{1 - \tan^{2} θ}{1 + \tan^{2} θ}$ . Therefore, on substituting value of $θ$ we get

$\Rightarrow \cos 30^{\circ} = \frac{1 - \tan^{2} 15^{\circ}}{1 + \tan^{2} 15^{\circ}}$

Substitute value of $\tan 30^{\circ} = \frac{\sqrt{3}}{2}$ .

$\Rightarrow \frac{\sqrt{3}}{2} = \frac{1 - \tan^{2} 15^{\circ}}{1 + \tan^{2} 15^{\circ}}$

$\Rightarrow \frac{1 - \tan^{2} 15^{\circ}}{1 + \tan^{2} 15^{\circ}} = \frac{\sqrt{3}}{2}$

Hence, option (c) is the correct answer.

The value of $\cos 1^{\circ} \cos 2^{\circ} \cos 3^{\circ} . . . . . \cos 179^{\circ}$ is

$\frac{1}{\sqrt{2}}$
$0$
$1$
$- 1$

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.

If $\tan θ = 3$ and $θ$ lies in third quadrant, then the value of $\sin θ$ is

$\frac{1}{\sqrt{10}}$
$- \frac{1}{\sqrt{10}}$
$\frac{- 3}{\sqrt{10}}$
$\frac{3}{\sqrt{10}}$

Ans: Given, $\tan θ = 3$

$\Rightarrow \tan θ = \frac{P}{B} = \frac{3}{1}$

$\Rightarrow H = \sqrt{3^{2} + 1^{2}}$

$\Rightarrow H = \sqrt{9 + 1}$

$\Rightarrow H = \sqrt{10}$

$\Rightarrow \sin θ = - \frac{3}{\sqrt{10}}$

( The value of $\sin θ$ is negative because $θ$ lies in the third quadrant )

Hence, option (a) is the correct answer.

The value of $\tan 75^{\circ} - \cot 75^{\circ}$ is equal to

$2 \sqrt{3}$
$2 + \sqrt{3}$
$2 - \sqrt{3}$
$1$

Ans: Given expression, $\tan 75^{\circ} - \cot 75^{\circ}$

Above written expression can also be written as,

$\Rightarrow \tan 75^{\circ} - \cot (90^{\circ} - 15^{\circ})$

We know that $\cot (90^{\circ} - θ) = \tan θ$ . Therefore, we get

$\Rightarrow \tan 75^{\circ} - \tan 15^{\circ}$

$\Rightarrow \frac{\sin 75^{\circ}}{\cos 75^{\circ}} - \frac{\sin 15^{\circ}}{\cos 15^{\circ}}$

Take LCM

$\Rightarrow \frac{\sin 75^{\circ} \cos 15^{\circ} - \cos 75^{\circ} \sin 15^{\circ}}{\cos 75^{\circ} \cos 15^{\circ}}$

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

$\Rightarrow \frac{\sin (75^{\circ} - 15^{\circ})}{\cos 75^{\circ} \cos 15^{\circ}}$

$\Rightarrow \frac{\sin (75^{\circ} - 15^{\circ})}{\frac{1}{2} \times 2 \cos 75^{\circ} \cos 15^{\circ}}$

We know that $2 \cos A \cos B = \cos (A + B) + \cos (A - B)$ . Therefore, we get

$\Rightarrow \frac{2 \sin (75^{\circ} - 15^{\circ})}{\cos (75^{\circ} + 15^{\circ}) + \cos (75^{\circ} - 15^{\circ})}$

$\Rightarrow \frac{2 \sin 60^{\circ}}{\cos 90^{\circ} + \cos 60^{\circ}}$

On substituting the values, we get

$\Rightarrow \frac{2 \times \frac{\sqrt{3}}{2}}{0 + \frac{1}{2}}$

$\Rightarrow \frac{\sqrt{3}}{0 + \frac{1}{2}} = 2 \sqrt{3}$

Hence, option (a) is the correct answer.

Which of the following is correct?

$\sin 1^{\circ} > \sin 1$
$\sin 1^{\circ} < \sin 1$
$\sin 1^{\circ} = \sin 1$
$\sin 1^{\circ} = \frac{π}{18^{\circ}} \sin 1$

Ans: We know that when $θ$ increases, the value of $\sin θ$ also increases.

Therefore, $\sin 1^{\circ} < \sin 1$

(Since $1 r a d i a n = \frac{π}{180} \sin 1$ )

Hence, option (d) is the correct answer.

If $\tan α = \frac{m}{m + 1}$ , $\tan β = \frac{1}{2 m + 1}$ , then $α + β$ is equal to

$\frac{π}{2}$
$\frac{π}{3}$
$\frac{π}{6}$
$\frac{π}{4}$

Ans: We have, $\tan α = \frac{m}{m + 1}$ and $\tan β = \frac{1}{2 m + 1}$

We know that $\tan (A + B) = \frac{\tan A + \tan B}{1 - \tan A . \tan B}$ . Therefore on substituting the values, we get

$\Rightarrow \tan (α + β) = \frac{\frac{m}{m + 1} + \frac{1}{2 m + 1}}{1 - \frac{m}{m + 1} . \frac{1}{2 m + 1}}$

$\Rightarrow \tan (α + β) = \frac{\frac{m (2 m + 1) + m + 1}{(m + 1) (2 m + 1)}}{\frac{(m + 1) (2 m + 1) - m}{(m + 1) (2 m + 1)}}$

On canceling common terms, we get

$\Rightarrow \tan (α + β) = \frac{2 m^{2} + m + m + 1}{2 m^{2} + m + 2 m + 1 - m}$

On simplification, we get

$\Rightarrow \tan (α + β) = \frac{2 m^{2} + 2 m + 1}{2 m^{2} + 2 m + 1} = 1$

We know that $\tan \frac{π}{4} = 1$ . Therefore, we get

$\Rightarrow \tan (α + β) = \tan \frac{π}{4}$

$\Rightarrow α + β = \frac{π}{4}$

Hence, option (d) is the correct answer.

The minimum value of $3 \cos x + 4 \sin x + 8$ is

Ans: Given expression, $3 \cos x + 4 \sin x + 8$

Let $y = 3 \cos x + 4 \sin x + 8$

Transport $8$ to the LHS

$\Rightarrow y - 8 = 3 \cos x + 4 \sin x . . . . . . (i)$

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.

The value of $\tan 3 A - \tan 2 A - \tan A$ is equal to

$\tan 3 A \tan 2 A \tan A$
$- \tan 3 A \tan 2 A \tan A$
$\tan A \tan 2 A - \tan 2 A \tan 3 A - \tan 3 A \tan A$
None of these

Ans: Given expression, $\tan 3 A - \tan 2 A - \tan A$

We can write $\tan 3 A = \tan (2 A + A)$ .

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

$\Rightarrow \tan 3 A = \frac{\tan 2 A + \tan A}{1 - \tan 2 A \tan A}$

Now, we will cross multiply the above written equation.

$\Rightarrow \tan 3 A (1 - \tan 2 A \tan A) = \tan 2 A + \tan A$

On multiplication of terms, we get

$\Rightarrow \tan 3 A - \tan 3 A \tan 2 A \tan A = \tan 2 A + \tan A$

$\Rightarrow \tan 3 A - \tan 2 A - \tan A = \tan 3 A \tan 2 A \tan A$

Hence, option (a) is the correct answer.

The value of $\sin (45^{\circ} + θ) - \cos (45^{\circ} - θ)$ is

$2 \cos θ$
$2 \sin θ$
$1$
$0$

Ans: Given, $\sin (45^{\circ} + θ) - \cos (45^{\circ} - θ)$

As we know $\sin (A + B) = \sin A \cos B + \cos A \sin B$ and $\cos (A - B) = \cos A \cos B + \sin A \sin B$ . Therefore, we get

$\Rightarrow (\sin 45^{\circ} \cos θ + \cos 45^{\circ} \sin θ) - (\cos 45^{\circ} \cos θ + \sin 45^{\circ} \sin θ)$

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

$\Rightarrow (\frac{1}{\sqrt{2}} \cos θ + \frac{1}{\sqrt{2}} \sin θ) - (\frac{1}{\sqrt{2}} \cos θ + \frac{1}{\sqrt{2}} \sin θ)$

$\Rightarrow \frac{1}{\sqrt{2}} \cos θ + \frac{1}{\sqrt{2}} \sin θ - \frac{1}{\sqrt{2}} \cos θ - \frac{1}{\sqrt{2}} \sin θ$

$\Rightarrow 0$

Hence, option (d) is the correct answer.

The value of $\cot (\frac{π}{4} + θ) \cot (\frac{π}{4} - θ)$ is

$- 1$
$0$
$1$
Not defined

Ans: We have, $\cot (\frac{π}{4} + θ) \cot (\frac{π}{4} - θ)$

We know that $\cot (x + y) = \frac{\cot x . \cot y - 1}{\cot y + \cot x}$ and $\cot (x - y) = \frac{\cot x . \cot y + 1}{\cot y - \cot x}$ . Therefore, we get

$\Rightarrow \frac{\cot \frac{π}{4} . \cot θ - 1}{\cot θ + \cot \frac{π}{4}} \times \frac{\cot \frac{π}{4} . \cot θ + 1}{\cot θ - \cot \frac{π}{4}}$

Substitute value of $\cos \frac{π}{4} = 1$ .

$\Rightarrow \frac{1. \cot θ - 1}{\cot θ + 1} \times \frac{1. \cot θ + 1}{\cot θ - 1}$

$\Rightarrow \frac{\cot θ - 1}{\cot θ + 1} \times \frac{\cot θ + 1}{\cot θ - 1}$

On canceling common terms, we get

$\Rightarrow 1$

Hence, option (c) is the correct answer.

$\cos 2 θ \cos 2 ϕ + \sin^{2} (θ - ϕ) - \sin^{2} (θ + ϕ)$ is equal to

$\sin 2 (θ + ϕ)$
$\cos 2 (θ + ϕ)$
$\sin 2 (θ - ϕ)$
$\cos 2 (θ - ϕ)$

Ans: We have, $\cos 2 θ \cos 2 ϕ + \sin^{2} (θ - ϕ) - \sin^{2} (θ + ϕ)$

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

$\Rightarrow \cos 2 θ \cos 2 ϕ + \sin (θ - ϕ + θ + ϕ) . \sin (θ - ϕ - θ - ϕ)$

$\Rightarrow \cos 2 θ \cos 2 ϕ + \sin (2 θ) . \sin (- 2 ϕ)$

We know that $\sin (- θ) = - \sin θ$ . Therefore, we get

$\Rightarrow \cos 2 θ \cos 2 ϕ - \sin 2 θ \sin 2 ϕ$

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

$\Rightarrow \cos 2 (θ + ϕ)$

Hence, option (b) is the correct answer.

The value of $\cos 12^{\circ} + \cos 84^{\circ} + \cos 156^{\circ} + \cos 132^{\circ}$ is

$\frac{1}{2}$
$1$
$- \frac{1}{2}$
$\frac{1}{8}$

Ans: Given expression: $\cos 12^{\circ} + \cos 84^{\circ} + \cos 156^{\circ} + \cos 132^{\circ}$

$\Rightarrow (\cos 132^{\circ} + \cos 12^{\circ}) + (\cos 156^{\circ} + \cos 84^{\circ})$

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

$\Rightarrow 2 \cos (\frac{132^{\circ} + 12^{\circ}}{2}) . \cos (\frac{132^{\circ} - 12^{\circ}}{2}) + 2 \cos (\frac{156^{\circ} + 84^{\circ}}{2}) . \cos (\frac{156^{\circ} - 84^{\circ}}{2})$

$\Rightarrow 2 \cos (\frac{144^{\circ}}{2}) . \cos (\frac{120^{\circ}}{2}) + 2 \cos (\frac{240^{\circ}}{2}) . \cos (\frac{72^{\circ}}{2})$

$\Rightarrow 2 \cos (72^{\circ}) . \cos (60^{\circ}) + 2 \cos (120^{\circ}) . \cos (36^{\circ})$

Substitute value of $\cos 60^{\circ} = \frac{1}{2}$ and $\cos 120^{\circ} = - \frac{1}{2}$ .

$\Rightarrow 2 \cos (72^{\circ}) \times \frac{1}{2} \times + 2 \times - \frac{1}{2} \times \cos (36^{\circ})$

$\Rightarrow \cos 72^{\circ} - \cos 36^{\circ}$

$\Rightarrow \cos (90^{\circ} - 18^{\circ}) - \cos 36^{\circ}$

We know that $\cos (90^{\circ} - θ) = \sin θ$ . Therefore, we get

$\Rightarrow \sin 18^{\circ} - \cos 36^{\circ}$

Substitute value of $\sin 18^{\circ} = \frac{\sqrt{5} - 1}{4}$ and $\cos 36^{\circ} = \frac{\sqrt{5} + 1}{4}$ .

$\Rightarrow \frac{\sqrt{5} - 1}{4} - \frac{\sqrt{5} + 1}{4}$

$\Rightarrow \frac{\sqrt{5} - 1 - \sqrt{5} - 1}{4}$

$\Rightarrow \frac{- 2}{4}$

$\Rightarrow \frac{- 1}{2}$

Hence, option (c) is the correct answer.

If $\tan A = \frac{1}{2}$ , $\tan B = \frac{1}{3}$ , then $\tan (2 A + B)$ is equal to

Ans: Given, $\tan A = \frac{1}{2}$ , $\tan B = \frac{1}{3}$

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

$\Rightarrow \tan 2 A = \frac{2 \tan A}{1 - \tan^{2} A}$

Now, we will substitute the value of $\tan A = \frac{1}{2}$

$\Rightarrow \tan 2 A = \frac{2 \times \frac{1}{2}}{1 - \frac{1^{2}}{2^{2}}}$

$\Rightarrow \tan 2 A = \frac{1}{\frac{4 - 1}{4}} = \frac{1}{\frac{3}{4}}$

$\Rightarrow \tan 2 A = \frac{4}{3}$

We need to find value of $\tan (2 A + B)$

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

$\Rightarrow \tan (2 A + B) = \frac{\tan 2 A + \tan B}{1 - \tan 2 A . \tan B}$

$\Rightarrow \tan (2 A + B) = \frac{\frac{4}{3} + \frac{1}{3}}{1 - \frac{4}{3} . \frac{1}{3}}$

$\Rightarrow \tan (2 A + B) = \frac{\frac{5}{3}}{\frac{9 - 4}{9}}$

$\Rightarrow \tan (2 A + B) = \frac{5}{3} \times \frac{9}{5}$

$\Rightarrow \tan (2 A + B) = 3$

Hence, option (c) is the correct answer.

The value of $\sin \frac{π}{10} \sin \frac{13 π}{10}$ is

$\frac{1}{2}$
$- \frac{1}{2}$
$- \frac{1}{4}$
$1$

Ans: We have, $\sin \frac{π}{10} \sin \frac{13 π}{10}$

$\Rightarrow \sin \frac{π}{10} \sin (π + \frac{3 π}{10})$

We know that $\sin (π + θ) = - \sin θ$ . Therefore, we get

$\Rightarrow \sin \frac{π}{10} \sin (- \frac{3 π}{10})$

We know that $\sin (- θ) = - \sin θ$ . Therefore, we get

$\Rightarrow - \sin \frac{180^{\circ}}{10} \sin (\frac{3 \times 180^{\circ}}{10})$

$\Rightarrow - \sin 18^{\circ} . \sin 54^{\circ}$

$\Rightarrow - \sin 18^{\circ} . \sin (90^{\circ} - 36^{\circ})$

We know that $\sin (90^{\circ} - θ) = \cos θ$ . Therefore, we get

$\Rightarrow - \sin 18^{\circ} . \cos 36^{\circ}$

On substituting the values $\sin 18^{\circ} = \frac{\sqrt{5} - 1}{4}$ and $\cos 36^{\circ} = \frac{\sqrt{5} + 1}{4}$ , we get

$\Rightarrow - (\frac{\sqrt{5} - 1}{4}) (\frac{\sqrt{5} + 1}{4})$

$\Rightarrow - (\frac{5 - 1}{16})$

$\Rightarrow - (\frac{4}{16}) = - \frac{1}{4}$

Hence, option (c) is the correct answer.

The value of $\sin 50^{\circ} - \sin 70^{\circ} + \sin 10^{\circ}$ is equal to

$1$
$0$
$\frac{1}{2}$
$2$

Ans: Given expression, $\sin 50^{\circ} - \sin 70^{\circ} + \sin 10^{\circ}$

$\Rightarrow (\sin 50^{\circ} - \sin 70^{\circ}) + \sin 10^{\circ}$

We know that $\sin C - \sin D = 2 \cos \frac{C + D}{2} \sin \frac{C - D}{2}$ . Therefore, we get

$\Rightarrow 2 \cos \frac{50^{\circ} + 70^{\circ}}{2} \sin \frac{50^{\circ} - 70^{\circ}}{2} + \sin 10^{\circ}$

$\Rightarrow 2 \cos 60^{\circ} \sin (- 10^{\circ}) + \sin 10^{\circ}$

We know that $\sin (- θ) = - \sin θ$ . Therefore, we get

$\Rightarrow - 2 \cos 60^{\circ} \sin 10^{\circ} + \sin 10^{\circ}$

Substitute value of $\sin 60^{\circ} = \frac{1}{2}$ .

$\Rightarrow - 2 \times \frac{1}{2} \times \sin 10^{\circ} + \sin 10^{\circ}$

$\Rightarrow - \sin 10^{\circ} + \sin 10^{\circ}$

$\Rightarrow 0$

Hence, option (b) is the correct answer.

If $\sin θ + \cos θ = 1$ , then the value of $\sin 2 θ$ is equal to

$1$
$\frac{1}{2}$
$0$
$- 1$

Ans: Given, $\sin θ + \cos θ = 1$

On squaring both the sides, we get

$\Rightarrow {(\sin θ + \cos θ)}^{2} = 1^{2}$

As we know ${(a + b)}^{2} = a^{2} + b^{2} + 2 a b$ . So, on applying this identity, we get

$\Rightarrow \sin^{2} θ + \cos^{2} θ + 2 \sin θ \cos θ = 1$

As we know $\sin^{2} θ + \cos^{2} θ = 1$ . So, on applying this formula, we get

$\Rightarrow 1 + 2 \sin θ \cos θ = 1$

As we know $\sin 2 A = 2 \sin A \cos A$ . Therefore, we get

$\Rightarrow 2 \sin θ \cos θ = 1 - 1$

$\Rightarrow \sin 2 θ = 0$

Hence, option (c) is the correct answer.

If $α + β = \frac{π}{4}$ , then the value of $(1 + \tan α) (1 + \tan β)$ is

$1$
$2$
$- 2$
Not defined

Ans: Given, $α + β = \frac{π}{4}$

$\Rightarrow \tan (α + β) = \tan \frac{π}{4}$

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

$\Rightarrow \frac{\tan α + \tan β}{1 - \tan α \tan β} = \tan \frac{π}{4}$

On substituting the value of $\tan \frac{π}{4} = 1$ , we get

$\Rightarrow \frac{\tan α + \tan β}{1 - \tan α \tan β} = 1$

On cross-multiplication, we get

$\Rightarrow \tan α + \tan β = 1 - \tan α \tan β$

$\Rightarrow \tan α + \tan β + \tan α \tan β = 1$

Add $1$ to both the sides

$\Rightarrow 1 + \tan α + \tan β + \tan α \tan β = 1 + 1$

Take $\tan β$ as a common term.

$\Rightarrow (1 + \tan α) + \tan β (1 + \tan α) = 2$

$\Rightarrow (1 + \tan α) (1 + \tan α) = 2$

Hence, option (b) is the correct answer.

If $\sin θ = \frac{- 4}{5}$ and $θ$ lies in third quadrant then the value of $\cos \frac{θ}{2}$ is

$\frac{1}{5}$
$- \frac{1}{\sqrt{10}}$
$- \frac{1}{\sqrt{5}}$
$\frac{1}{\sqrt{10}}$

Ans: Given, $\sin θ = \frac{- 4}{5}$ . Here, value of $\sin e$ is negative because $θ$ lies in third quadrant. As here we need to find $\cos \frac{θ}{2}$ , so we will first find the value of $\cos θ$ .

We know that $\sin^{2} θ + \cos^{2} θ = 1$ .

$\Rightarrow \cos θ = \sqrt{1 - \sin^{2} θ}$

Substitute the value of $\sin θ = \frac{- 4}{5}$ .

$\Rightarrow \cos θ = \sqrt{1 - {(\frac{- 4}{5})}^{2}}$

$\Rightarrow \cos θ = \sqrt{1 - \frac{16}{25}}$

On taking LCM, we get

$\Rightarrow \cos θ = \sqrt{\frac{25 - 16}{25}}$

$\Rightarrow \cos θ = \sqrt{\frac{9}{25}}$

$\Rightarrow \cos θ = \pm \frac{3}{5}$

As it is mentioned in the question that $θ$ lies in the third quadrant and we know $\cos$ is negative in the third quadrant. Therefore, we get

$\Rightarrow \cos θ = - \frac{3}{5}$

We know that $\cos θ = 2 \cos^{2} \frac{θ}{2} - 1$ . Now we will substitute value of $\cos θ$ in this identity.

$\Rightarrow \frac{- 3}{5} = 2 \cos^{2} \frac{θ}{2} - 1$

(Here, value of $\cos θ$ is negative because $π < θ < \frac{3 π}{2}$ )

$\Rightarrow 2 \cos^{2} \frac{θ}{2} = 1 - \frac{3}{5}$

$\Rightarrow 2 \cos^{2} \frac{θ}{2} = \frac{2}{5}$

On canceling common term, we get

$\Rightarrow \cos^{2} \frac{θ}{2} = \frac{1}{5}$

$\Rightarrow \cos \frac{θ}{2} = \pm \frac{1}{\sqrt{5}}$

As we know $π < θ < \frac{3 π}{2}$ . Therefore, $\frac{π}{2} < \frac{θ}{2} < \frac{3 π}{4}$

$\Rightarrow \cos \frac{θ}{2} = - \frac{1}{\sqrt{5}}$

Hence, option (c) is the correct answer.

Number of solutions of the equation $\tan x + \sec x = 2 \cos x$ lying in the interval $[0, 2 π]$ is

Ans: Given that, $\tan x + \sec x = 2 \cos x$

We will first write above written equation in terms of $\sin e$ and $\cos$ .

$\Rightarrow \frac{\sin x}{\cos x} + \frac{1}{\cos x} = 2 \cos x$

$\Rightarrow \frac{\sin x + 1}{\cos x} = 2 \cos x$

$\Rightarrow \sin x + 1 = 2 \cos^{2} x$

$\Rightarrow 2 \cos^{2} x - \sin x - 1 = 0$

$\Rightarrow 2 (1 - \sin^{2} x) - \sin x - 1 = 0$

$\Rightarrow - 2 \sin^{2} x - \sin x + 1 = 0$

$\Rightarrow 2 \sin^{2} x + \sin x - 1 = 0$

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.

The value of $\sin \frac{π}{18} + \sin \frac{π}{9} + \sin \frac{2 π}{9} + \sin \frac{5 π}{18}$ is given by

$\sin \frac{7 π}{18} + \sin \frac{4 π}{9}$
$1$
$\cos \frac{π}{6} + \cos \frac{3 π}{7}$
$\cos \frac{π}{9} + \sin \frac{π}{9}$

Ans: Given expression: $\sin \frac{π}{18} + \sin \frac{π}{9} + \sin \frac{2 π}{9} + \sin \frac{5 π}{18}$

$\Rightarrow (\sin \frac{5 π}{18} + \sin \frac{π}{18}) + (\sin \frac{2 π}{9} + \sin \frac{π}{9})$

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

$\Rightarrow 2 \sin (\frac{\frac{5 π}{18} + \frac{π}{18}}{2}) . \cos (\frac{\frac{5 π}{18} - \frac{π}{18}}{2}) + 2 \sin (\frac{\frac{2 π}{9} + \frac{π}{9}}{2}) . \cos (\frac{\frac{2 π}{9} - \frac{π}{9}}{2})$

$\Rightarrow 2 \sin (\frac{\frac{6 π}{18}}{2}) . \cos (\frac{\frac{4 π}{18}}{2}) + 2 \sin (\frac{\frac{3 π}{9}}{2}) . \cos (\frac{\frac{π}{9}}{2})$

$\Rightarrow 2 \sin (\frac{π}{6}) . \cos (\frac{π}{9}) + 2 \sin (\frac{π}{6}) . \cos (\frac{π}{18})$

Substitute value of $\sin \frac{π}{6} = \frac{1}{2}$ .

$\Rightarrow 2 \times \frac{1}{2} \times \cos (\frac{π}{9}) + 2 \times \frac{1}{2} \times \cos (\frac{π}{18})$

$\Rightarrow \cos (\frac{π}{9}) + \cos (\frac{π}{18})$

We know that $\cos θ = \sin (90^{\circ} - θ)$ . Therefore, we get

$\Rightarrow \sin (\frac{π}{2} - \frac{π}{9}) + \sin (\frac{π}{2} - \frac{π}{18})$

$\Rightarrow \sin \frac{7 π}{18} + \sin \frac{8 π}{18}$

$\Rightarrow \sin \frac{7 π}{18} + \sin \frac{4 π}{9}$

Hence, option (a) is the correct answer.

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

$\frac{- 53}{10}$
$\frac{23}{10}$
$\frac{37}{10}$
$\frac{7}{10}$

Ans: Given, $3 \tan A + 4 = 0$

$\Rightarrow 3 \tan A = - 4$

$\Rightarrow \tan A = \frac{- 4}{3} = \frac{P}{B}$

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

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

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

$\Rightarrow H^{2} = 25$

$\Rightarrow H = 5$

We know that $\cos x = \frac{B}{H}$ . Therefore, we get

$\Rightarrow \cos A = \frac{- 3}{5}$ (Negative because $A$ lies in second quadrant)

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

$\Rightarrow \sin A = \frac{4}{5}$

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

$\Rightarrow \cot A = \frac{- 3}{4}$ (Negative because $A$ lies in second quadrant)

We need to find the value of $2 \cot A - 5 \cos A + \sin A$

$\Rightarrow 2 \times \frac{- 3}{4} - 5 \times (\frac{- 3}{5}) + \frac{4}{5}$

$\Rightarrow \frac{- 3}{2} + 3 + \frac{4}{5}$

$\Rightarrow \frac{- 15 + 30 + 8}{10}$

$\Rightarrow \frac{23}{10}$

Hence, option (a) is the correct answer.

The value of $\cos^{2} 48^{\circ} - \sin^{2} 12^{\circ}$ is

$\frac{\sqrt{5} + 1}{8}$
$\frac{\sqrt{5} - 1}{8}$
$\frac{\sqrt{5} + 1}{5}$
$\frac{\sqrt{5} + 1}{2 \sqrt{2}}$

Ans: Given expression: $\cos^{2} 48^{\circ} - \sin^{2} 12^{\circ}$

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

$\Rightarrow \cos (48^{\circ} + 12^{\circ}) . \cos (48^{\circ} - 12^{\circ})$

$\Rightarrow \cos 60^{\circ} . \cos 36^{\circ}$

On substituting the values, $\cos 60^{\circ} = \frac{1}{2}$ and $\cos 36^{\circ} = \frac{\sqrt{5} + 1}{4}$ . Therefore, we get

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

$\Rightarrow \frac{\sqrt{5} + 1}{8}$

Hence, option (a) is the correct answer.

If $\tan α = \frac{1}{7}$ , $\tan β = \frac{1}{3}$ , then $\cos 2 α$ is equal to

$\sin 2 β$
$\sin 4 β$
$\sin 3 β$
$\cos 2 β$

Ans: Given, $\tan α = \frac{1}{7}$ , $\tan β = \frac{1}{3}$

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

$\Rightarrow \cos 2 α = \frac{1 - \tan^{2} α}{1 + \tan^{2} α} = \frac{1 - \frac{1^{2}}{7^{2}}}{1 + \frac{1^{2}}{7^{2}}}$

$\Rightarrow \cos 2 α = \frac{\frac{7^{2} - 1}{7^{2}}}{\frac{7^{2} + 1}{7^{2}}} = \frac{7^{2} - 1}{7^{2} + 1}$

$\Rightarrow \cos 2 α = \frac{49 - 1}{49 + 1} = \frac{48}{50}$

$\Rightarrow \cos 2 α = \frac{24}{25}$

Similarly,

$\Rightarrow \tan 2 β = \frac{2 \tan β}{1 - \tan^{2} β}$

$\Rightarrow \tan 2 β = \frac{2 \times \frac{1}{3}}{1 - \frac{1^{2}}{3^{2}}}$

$\Rightarrow \tan 2 β = \frac{\frac{2}{3}}{\frac{8}{9}} = \frac{2}{3} \times \frac{9}{8}$

$\Rightarrow \tan 2 β = \frac{3}{4}$

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

$\Rightarrow \sin 4 β = \frac{2 \tan 2 β}{1 + \tan^{2} 2 β}$

$\Rightarrow \sin 4 β = \frac{2 \times \frac{3}{4}}{1 + \frac{3^{2}}{4^{2}}}$

$\Rightarrow \sin 4 β = \frac{\frac{6}{4}}{\frac{16 + 9}{16}}$

$\Rightarrow \sin 4 β = \frac{6}{4} \times \frac{16}{25}$

$\Rightarrow \sin 4 β = \frac{24}{25} = \cos 2 α$

Hence, option (b) is the correct answer.

If $\tan θ = \frac{a}{b}$ , then $b \cos 2 θ + a \sin 2 θ$ is equal to

$a$
$b$
$\frac{a}{b}$
None

Ans: Given, $\tan θ = \frac{a}{b}$

We have, $b \cos 2 θ + a \sin 2 θ$

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

$\Rightarrow b [\frac{1 - \tan^{2} θ}{1 + \tan^{2} θ}] + a [\frac{2 \tan θ}{1 + \tan^{2} θ}]$

Substitute $\tan θ = \frac{a}{b}$

$\Rightarrow b [\frac{1 - \frac{a^{2}}{b^{2}}}{1 + \frac{a^{2}}{b^{2}}}] + a [\frac{2 \frac{a}{b}}{1 + \frac{a^{2}}{b^{2}}}]$

$\Rightarrow b [\frac{\frac{b^{2} - a^{2}}{b^{2}}}{\frac{b^{2} + a^{2}}{b^{2}}}] + a [\frac{2 \frac{a}{b}}{\frac{b^{2} + a^{2}}{b^{2}}}]$

$\Rightarrow b [\frac{b^{2} - a^{2}}{b^{2} + a^{2}}] + a [\frac{2 a b}{b^{2} + a^{2}}]$

$\Rightarrow [\frac{b^{3} - a^{2} b}{b^{2} + a^{2}}] + [\frac{2 a^{2} b}{b^{2} + a^{2}}]$

$\Rightarrow \frac{b^{3} - a^{2} b + 2 a^{2} b}{b^{2} + a^{2}}$

$\Rightarrow \frac{b (b^{2} + a^{2})}{b^{2} + a^{2}}$

$\Rightarrow b$

Hence, option (b) is the correct answer.

If for real values of $x$ , $\cos θ = x + \frac{1}{x}$ , then

$θ$ is an acute angle
$θ$ is right angle
$θ$ is an obtuse angle
No value of $θ$ is possible

Ans: Given, $\cos θ = x + \frac{1}{x}$

$\Rightarrow \cos θ = \frac{x^{2} + 1}{x}$

$\Rightarrow x^{2} + 1 = x \cos θ$

$\Rightarrow x^{2} - x \cos θ + 1 = 0$

We know that for real value of $x$ , $b^{2} - 4 a c ⩾ 0$ . Therefore, we get

$\Rightarrow {(- \cos θ)}^{2} - 4 \times 1 \times 1 ⩾ 0$

$\Rightarrow {(- \cos θ)}^{2} - 4 ⩾ 0$

$\Rightarrow \cos^{2} θ ⩾ 4$

$\Rightarrow \cos θ ⩾ \pm 2$

We know that $- 1 ⩽ \cos θ ⩽ 1$ . Therefore, value of $θ$ is not possible

Hence, option (d) is the correct answer.

Fill in the blanks in exercises $60$ to $67$ .

The value of $\frac{\sin 50^{\circ}}{\sin 130^{\circ}}$ is …….

Ans: We need to find the value of $\frac{\sin 50^{\circ}}{\sin 130^{\circ}}$

As we know $\sin (180^{\circ} - θ) = \sin θ$ . Therefore, we get

$\Rightarrow \frac{\sin 50^{\circ}}{\sin (180^{\circ} - 50^{\circ})}$

$\Rightarrow \frac{\sin 50^{\circ}}{\sin 50^{\circ}}$

On canceling the common term, we get

$\Rightarrow 1$

Thus, value of filler is $1$ .

If $k = \sin (\frac{π}{18}) \sin (\frac{5 π}{18}) \sin (\frac{7 π}{18})$ , then the numerical value of $k$ is …….

Ans: Given, $k = \sin (\frac{π}{18}) \sin (\frac{5 π}{18}) \sin (\frac{7 π}{18})$

Substitute value of $π = 180^{\circ}$ .

$\Rightarrow k = \sin (\frac{180^{\circ}}{18}) \sin (\frac{5 \times 180^{\circ}}{18}) \sin (\frac{7 \times 180^{\circ}}{18})$

On simplification, we get

$\Rightarrow k = \sin 10^{\circ} . \sin 50^{\circ} . \sin 70^{\circ}$

$\Rightarrow k = \sin 10^{\circ} . \sin (90^{\circ} - 40^{\circ}) . \sin (90^{\circ} - 20^{\circ})$

As we know $\sin (90^{\circ} - θ) = \cos θ$ . Therefore, we get

$\Rightarrow k = \sin 10^{\circ} . \cos 40^{\circ} . \cos 20^{\circ}$

Multiply and divide above written equation by $2$

$\Rightarrow k = \sin 10^{\circ} \frac{1}{2} [2 \cos 40^{\circ} \cos 20^{\circ}]$

We know that $2 \cos x \cos y = \cos (x + y) + \cos (x - y)$ . Therefore, we get

$\Rightarrow k = \frac{1}{2} \sin 10^{\circ} [\cos (40^{\circ} + 20^{\circ}) + \cos (40^{\circ} - 20^{\circ})]$

$\Rightarrow k = \frac{1}{2} \sin 10^{\circ} [\cos 60^{\circ} + \cos 20^{\circ}]$

Substitute value of $\cos 60^{\circ} = \frac{1}{2}$ .

$\Rightarrow k = \frac{1}{2} \sin 10^{\circ} [\frac{1}{2} + \cos 20^{\circ}]$

On multiplication of terms, we get

$\Rightarrow k = \frac{1}{4} \sin 10^{\circ} + \frac{1}{2} \sin 10^{\circ} \cos 20^{\circ}$

Multiply and divide $\sin 10^{\circ} \cos 20^{\circ}$ by $2$

$\Rightarrow k = \frac{1}{4} \sin 10^{\circ} + \frac{1}{2} \times \frac{1}{2} (2 \sin 10^{\circ} \cos 20^{\circ})$

We know that $2 \sin x \cos y = \sin (x + y) + \sin (x - y)$ . Therefore, we get

$\Rightarrow k = \frac{1}{4} \sin 10^{\circ} + \frac{1}{4} (\sin (10^{\circ} + 20^{\circ}) + \sin (10^{\circ} - 20^{\circ}))$

$\Rightarrow k = \frac{1}{4} \sin 10^{\circ} + \frac{1}{4} (\sin 30^{\circ} + \sin (- 10^{\circ}))$

We know that $\sin (- θ) = - \sin θ$ . Therefore, we get

$\Rightarrow k = \frac{1}{4} \sin 10^{\circ} + \frac{1}{4} \sin 30^{\circ} - \frac{1}{4} \sin 10^{\circ}$

$\Rightarrow k = \frac{1}{4} \sin 30^{\circ}$

Substitute the value of $\sin 30^{\circ} = \frac{1}{2}$ .

$\Rightarrow k = \frac{1}{4} \times \frac{1}{2} = \frac{1}{8}$

Thus, value of filler is $\frac{1}{8}$ .

If $\tan A = \frac{1 - \cos B}{\sin B}$ , then $\tan 2 A =$ …….

Ans: Given, $\tan A = \frac{1 - \cos B}{\sin B}$

We know that $\tan 2 A = \frac{2 \tan A}{1 - \tan^{2} A}$ . So, we will substitute given value of $\tan A$ in this formula.

$\Rightarrow \tan 2 A = \frac{2 (\frac{1 - \cos B}{\sin B})}{1 - {(\frac{1 - \cos B}{\sin B})}^{2}}$

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

$\Rightarrow \tan 2 A = \frac{2 (\frac{2 \sin^{2} \frac{B}{2}}{2 \sin \frac{B}{2} \cos \frac{B}{2}})}{1 - {(\frac{2 \sin^{2} \frac{B}{2}}{2 \sin \frac{B}{2} \cos \frac{B}{2}})}^{2}}$

$\Rightarrow \tan 2 A = \frac{2 (\frac{\sin \frac{B}{2}}{\cos \frac{B}{2}})}{1 - {(\frac{\sin \frac{B}{2}}{\cos \frac{B}{2}})}^{2}}$

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

$\Rightarrow \tan 2 A = \frac{2 (\tan \frac{B}{2})}{1 - \tan^{2} \frac{B}{2}}$

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

$\Rightarrow \tan 2 A = \tan B$

Thus, value of filler is $\tan B$ .

If $\sin x + \cos x = a$ , then

$\sin^{6} x + \cos^{6} x =$ ……
$| \sin x - \cos x | =$ ……

Ans: Given, $\sin x + \cos x = a$

On squaring both the sides, we get

$\Rightarrow {(\sin x + \cos x)}^{2} = a^{2}$

$\Rightarrow \sin^{2} x + \cos^{2} x + 2 \sin x \cos x = a^{2}$

$\Rightarrow 1 + 2 \sin x \cos x = a^{2}$

$\Rightarrow 2 \sin x \cos x = a^{2} - 1$

$\Rightarrow \sin x \cos x = \frac{a^{2} - 1}{2} . . . . . (1)$

$\sin^{6} x + \cos^{6} x$

This can also be written as ${(\sin^{2} x)}^{3} + {(\cos^{2} x)}^{3}$ .

We know that $a^{3} + b^{3} = {(a + b)}^{3} - 3 a b (a + b)$ . Therefore, we get

$\Rightarrow {(\sin^{2} x + \cos^{2} x)}^{3} - 3 \sin^{2} x \cos^{2} x (\sin^{2} x + \cos^{2} x)$

$\Rightarrow 1 - 3 \sin^{2} x \cos^{2} x$

On substituting $\sin x \cos x = \frac{a^{2} - 1}{2}$ , we get

$\Rightarrow 1 - 3 {(\frac{a^{2} - 1}{2})}^{2}$

$\Rightarrow 1 - \frac{3 {(a^{2} - 1)}^{2}}{4}$

$\Rightarrow \frac{1}{4} [4 - 3 {(a^{2} - 1)}^{2}]$

$| \sin x - \cos x |$

On squaring, we get

$\Rightarrow {| \sin x - \cos x |}^{2}$

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

On substituting $\sin x \cos x = \frac{a^{2} - 1}{2}$ , we get

$\Rightarrow 1 - 2 (\frac{a^{2} - 1}{2})$

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

$\Rightarrow 2 - a^{2}$

$\Rightarrow | \sin x - \cos x | = \sqrt{2 - a^{2}}$

We know that $| \sin x - \cos x | > 0$ .

Therefore, the value of filler is $\sqrt{2 - a^{2}}$ .

In a triangle ABC with $∠ C = 90^{\circ}$ the equation whose roots are $\tan A$ and $\tan B$ is ……

Ans: Given, $△ A B C$ with $∠ C = 90^{\circ}$

We know that quadratic equation in the form of roots is $x^{2} - (α + β) x + α β = 0$ .

$\Rightarrow x^{2} - (\tan A + \tan B) x + \tan A . \tan B = 0$

We know that sum of interior angle of triangle is $180^{\circ}$ . Therefore, we get

$\Rightarrow ∠ A + ∠ B + ∠ C = 180^{\circ}$

$\Rightarrow ∠ A + ∠ B + 90^{\circ} = 180^{\circ}$

$\Rightarrow ∠ A + ∠ B = 180^{\circ}$

$\Rightarrow \tan (A + B) = \tan 90^{\circ}$

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

$\Rightarrow \frac{\tan A + \tan B}{1 - \tan A \tan B} = \frac{1}{0}$

On cross multiplication, we get

$\Rightarrow 1 - \tan A \tan B = 0$

$\Rightarrow \tan A \tan B = 1. . . . (i)$

Now, $\tan A + \tan B$

$\Rightarrow \frac{\sin A}{\cos A} + \frac{\sin B}{\cos B}$

$\Rightarrow \frac{\sin A \cos B + \cos A \sin B}{\cos A \cos B}$

We know that $\sin (A + B) = \sin A \cos B + \cos A \sin B$ . Therefore, we get

$\Rightarrow \frac{\sin (A + B)}{\cos A \cos B}$

$\Rightarrow \frac{\sin 90^{\circ}}{\cos A \cos (90^{\circ} - A)}$

$\Rightarrow \frac{1}{\cos A \sin A}$

$\Rightarrow \frac{2}{2 \cos A \sin A}$

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

$\Rightarrow \frac{2}{\sin 2 A}$

$\Rightarrow \tan A + \tan B = \frac{2}{\sin 2 A} . . . . . (i i)$

We have, $x^{2} - (\tan A + \tan B) x + \tan A . \tan B = 0$ . Therefore, from equation $(i)$ and $(i i)$ , we get

$\Rightarrow x^{2} - (\frac{2}{\sin 2 A}) x + 1 = 0$

Therefore, the value of filler is $x^{2} - (\frac{2}{\sin 2 A}) x + 1 = 0$ .

$3 {(\sin x - \cos x)}^{4} + 6 {(\sin x + \cos x)}^{2} + 4 (\sin^{6} x + \cos^{6} x) =$ …….

Ans: Given expression: $3 {(\sin x - \cos x)}^{4} + 6 {(\sin x + \cos x)}^{2} + 4 (\sin^{6} x + \cos^{6} x)$

$\Rightarrow 3 {[{(\sin x - \cos x)}^{2}]}^{2} + 6 {(\sin x + \cos x)}^{2} + 4 (\sin^{6} x + \cos^{6} x)$

$\Rightarrow 3 {[\sin^{2} x + \cos^{2} x - 2 \sin x \cos x]}^{2} + 6 [\sin^{2} x + \cos^{2} x + 2 \sin x \cos x] + 4 [{(\sin^{2} x)}^{3} + {(\cos^{2} x)}^{3}]$

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

$\Rightarrow 3 {[1 - 2 \sin x \cos x]}^{2} + 6 [1 + 2 \sin x \cos x] + 4 [{(\sin^{2} x)}^{3} + {(\cos^{2} x)}^{3}]$

We know that $a^{3} + b^{3} = {(a + b)}^{3} - 3 a b (a + b)$ . Therefore, we get

$\Rightarrow 3 {[1 - 2 \sin x \cos x]}^{2} + 6 [1 + 2 \sin x \cos x] + 4 [{(\sin^{2} x + \cos^{2} x)}^{3} - 3 \sin^{2} x \cos^{2} x (\sin^{2} x + \cos^{2} x)]$

$\Rightarrow 3 [1 + 4 \sin^{2} x \cos^{2} x - 4 \sin x \cos x] + 6 [1 + 2 \sin x \cos x] + 4 [1 - 3 \sin^{2} x \cos^{2} x]$

$\Rightarrow 3 + 12 \sin^{2} x \cos^{2} x - 12 \sin x \cos x + 6 + 12 \sin x \cos x + 4 - 12 \sin^{2} x \cos^{2} x$

$\Rightarrow 3 + 6 + 4$

$\Rightarrow 13$

Therefore, the value of filler is $13$ .

Given $x > 0$ , the values of $f (x) = - 3 \cos \sqrt{3 + x + x^{2}}$ lie in the interval …….

Ans: Given, $f (x) = - 3 \cos \sqrt{3 + x + x^{2}}$

Put $\sqrt{3 + x + x^{2}} = y$

$\Rightarrow f (x) = - 3 \cos y$

We know that $- 1 ⩽ \cos y ⩽ 1$ .

$\Rightarrow 3 ⩾ - 3 \cos y ⩾ - 3$

$\Rightarrow - 3 ⩽ - 3 \cos y ⩽ 3$

Substitute value of $y$ .

$\Rightarrow - 3 ⩽ - 3 \cos \sqrt{3 + x + x^{2}} ⩽ 3$

Therefore, the value of filler is $[- 3, 3]$ .

The maximum distance of a point on the graph of the function $y = \sqrt{3} \sin x + \cos x$ from $x - a x i s$ is …….

Ans: Given, $f (x) = - 3 \cos \sqrt{3 + x + x^{2}}$

$\Rightarrow y = - 3 \cos \sqrt{3 + x + x^{2}}$

We know that the maximum distance from a point on the graph from $x - a x i s$ is given by,

$\Rightarrow \sqrt{{(\sqrt{3})}^{2} + {(1)}^{2}}$

$\Rightarrow \sqrt{3 + 1}$

$\Rightarrow 2$

Therefore, the value of filler is $2$ .

In each of the exercises $68$ to $75$ , state whether the statements is True or False? Also give justification.

If $\tan A = \frac{1 - \cos B}{\sin B}$ , then $\tan 2 A = \tan B$ .

Ans: Given, $\tan A = \frac{1 - \cos B}{\sin B}$

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

$\Rightarrow \tan A = \frac{2 \sin^{2} \frac{B}{2}}{2 \sin \frac{B}{2} \cos \frac{B}{2}}$

$\Rightarrow \tan A = \frac{\sin \frac{B}{2}}{\cos \frac{B}{2}}$

$\Rightarrow \tan A = \tan \frac{B}{2}$

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

$\Rightarrow \tan 2 A = \frac{2 \tan A}{1 - \tan^{2} A}$

On substituting the value of $\tan A = \tan \frac{B}{2}$ , we get

$\Rightarrow \tan 2 A = \frac{2 \tan \frac{B}{2}}{1 - \tan^{2} \frac{B}{2}}$

$\Rightarrow \tan 2 A = \tan B$

Thus, the given statement is true.

The equality $\sin A + \sin 2 A + \sin 3 A = 3$ holds for some real value of $A$ .

Ans: Given, $\sin A + \sin 2 A + \sin 3 A = 3$

We know that the maximum value of $\sin A$ is $1$ . But for $\sin 2 A$ and $\sin 3 A$ it is not equal to $1$ . Hence, it is not possible

Thus, the given statement is false.

$\sin 10^{\circ}$ is greater than $\cos 10^{\circ}$ .

Ans: Given, $\sin 10^{\circ} > \cos 10^{\circ}$

$\Rightarrow \sin 10^{\circ} > \cos (90^{\circ} - 80^{\circ})$

We know that $\cos (90^{\circ} - θ) = \sin θ$ . Therefore, we get

$\Rightarrow \sin 10^{\circ} > \sin 80^{\circ}$

It is incorrect because value of $\sin e$ is in increasing order.

Thus, the given statement is false.

$\cos \frac{2 π}{15} \cos \frac{4 π}{15} \cos \frac{8 π}{15} \cos \frac{16 π}{15} = \frac{1}{16}$ .

Ans: Let us start to solve $\cos \frac{2 π}{15} \cos \frac{4 π}{15} \cos \frac{8 π}{15} \cos \frac{16 π}{15}$

Substitute value of $π = 180^{\circ}$ .

$\Rightarrow \cos \frac{2 \times 180^{\circ}}{15} . \cos \frac{4 \times 180^{\circ}}{15} . \cos \frac{8 \times 180^{\circ}}{15} . \cos \frac{16 \times 180^{\circ}}{15}$

$\Rightarrow \cos \frac{2 \times 12^{\circ}}{1} . \cos \frac{4 \times 12^{\circ}}{1} . \cos \frac{8 \times 12^{\circ}}{1} . \cos \frac{16 \times 12^{\circ}}{1}$

$\Rightarrow \cos 24^{\circ} . \cos 48^{\circ} . \cos 96^{\circ} . \cos 192^{\circ}$

Now, we will multiply and divide the above written expression by $16 \sin 24^{\circ}$ .

$\Rightarrow \frac{16 \sin 24^{\circ}}{16 \sin 24^{\circ}} [\cos 24^{\circ} . \cos 48^{\circ} . \cos 96^{\circ} . \cos 192^{\circ}]$

Above written expression can also be written as,

$\Rightarrow \frac{1}{16 \sin 24^{\circ}} [(2 \sin 24^{\circ} \cos 24^{\circ}) .2 \cos 48^{\circ} .2 \cos 96^{\circ} .2 \cos 192^{\circ}]$

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

$\Rightarrow \frac{1}{16 \sin 24^{\circ}} [\sin (2 \times 24^{\circ}) .2 \cos 48^{\circ} .2 \cos 96^{\circ} .2 \cos 192^{\circ}]$

$\Rightarrow \frac{1}{16 \sin 24^{\circ}} [\sin 48^{\circ} .2 \cos 48^{\circ} .2 \cos 96^{\circ} .2 \cos 192^{\circ}]$

$\Rightarrow \frac{1}{16 \sin 24^{\circ}} [(2 \sin 48^{\circ} \cos 48^{\circ}) .2 \cos 96^{\circ} .2 \cos 192^{\circ}]$

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

$\Rightarrow \frac{1}{16 \sin 24^{\circ}} [\sin 96^{\circ} .2 \cos 96^{\circ} .2 \cos 192^{\circ}]$

$\Rightarrow \frac{1}{16 \sin 24^{\circ}} [2 \sin 96^{\circ} \cos 96^{\circ} .2 \cos 192^{\circ}]$

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

$\Rightarrow \frac{1}{16 \sin 24^{\circ}} [\sin 192^{\circ} .2 \cos 192^{\circ}]$

$\Rightarrow \frac{1}{16 \sin 24^{\circ}} [2 \sin 192^{\circ} \cos 192^{\circ}]$

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

$\Rightarrow \frac{1}{16 \sin 24^{\circ}} (\sin 384^{\circ})$

We know that $\sin 384^{\circ}$ can also be written as $\sin (360^{\circ} + 24^{\circ})$ . Therefore, we get

$\Rightarrow \frac{1}{16 \sin 24^{\circ}} \sin (360^{\circ} + 24^{\circ})$

We know that $\sin (360^{\circ} + 24^{\circ}) = \sin 24^{\circ}$ . Therefore, we get

$\Rightarrow \frac{1}{16 \sin 24^{\circ}} \times \sin 24^{\circ}$

$\Rightarrow \frac{1}{16}$

Hence proved L.HS. = R.H.S.

Thus, the given statement is true.

One value of $θ$ which satisfies the equation $\sin^{4} θ - 2 \sin^{2} θ - 1$ lies between $0$ and $2 π$ .

Ans: We have, $\sin^{4} θ - 2 \sin^{2} θ - 1$

Let $y = \sin^{2} θ$ . Therefore, we get

$\Rightarrow y^{2} - 2 y - 1 = 0$

We know that for quadratic equation $a x^{2} + b x + c = 0$ , $x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ . Therefore, we get

$\Rightarrow y = \frac{- (- 2) \pm \sqrt{{(- 2)}^{2} - 4 \times 1 \times - 1}}{2 \times 1}$

$\Rightarrow y = \frac{2 \pm \sqrt{4 + 4}}{2}$

$\Rightarrow y = \frac{2 \pm 2 \sqrt{2}}{2} = \frac{2 (1 \pm \sqrt{2})}{2}$

On canceling common terms, we get

$\Rightarrow y = 1 \pm \sqrt{2}$

$\Rightarrow \sin^{2} θ = 1 \pm \sqrt{2}$

$\Rightarrow \sin^{2} θ = 1 + \sqrt{2}$ or $\sin^{2} θ = 1 - \sqrt{2}$

We know that $- 1 ⩽ \sin θ ⩽ 1$ . Therefore, we say that $\sin^{2} θ ⩽ 1$ .

But we have,

$\Rightarrow \sin^{2} θ = 1 + \sqrt{2}$ or $\sin^{2} θ = 1 - \sqrt{2}$

Which is not possible.

Thus, the given statement is false.

If $\cos e c x = 1 + \cot x$ then $x = 2 n π, 2 n π + \frac{π}{2}$ .

Ans: Given, $\cos e c x = 1 + \cot x$

$\Rightarrow \frac{1}{\sin x} = 1 + \frac{\cos x}{\sin x}$

$\Rightarrow \frac{1}{\sin x} = \frac{\sin x + \cos x}{\sin x}$

$\Rightarrow \sin x + \cos x = 1$

Divide whole equation by $\sqrt{2}$ .

$\Rightarrow \frac{1}{\sqrt{2}} \sin x + \frac{1}{\sqrt{2}} \cos x = \frac{1}{\sqrt{2}}$

We know that $\sin \frac{π}{4} = \frac{1}{\sqrt{2}}$ and $\cos \frac{π}{4} = \frac{1}{\sqrt{2}}$ . Therefore, we can write above written equation as,

$\Rightarrow \sin \frac{π}{4} \sin x + \cos \frac{π}{4} \cos x = \frac{1}{\sqrt{2}}$

$\Rightarrow \cos x \cos \frac{π}{4} + \sin x \sin \frac{π}{4} = \frac{1}{\sqrt{2}}$

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

$\Rightarrow \cos (x - \frac{π}{4}) = \frac{1}{\sqrt{2}}$

$\Rightarrow \cos (x - \frac{π}{4}) = \cos \frac{π}{4}$

We know that if $\cos θ = \cos α$ , then $θ = 2 n π \pm α$ . Therefore, we get

$\Rightarrow x - \frac{π}{4} = 2 n π \pm \frac{π}{4}$

$\Rightarrow x = 2 n π \pm \frac{π}{4} + \frac{π}{4}$

$\Rightarrow x = 2 n π + \frac{π}{4} + \frac{π}{4}$ or $\Rightarrow x = 2 n π - \frac{π}{4} + \frac{π}{4}$

$\Rightarrow x = 2 n π + \frac{π}{2}$ or $\Rightarrow x = 2 n π$

Thus, the given statement is true.

If $\tan θ + \tan 2 θ + \sqrt{3} \tan θ \tan 2 θ = \sqrt{3}$ , then $θ = \frac{n π}{3} + \frac{π}{9}$ .

Ans: Given, $\tan θ + \tan 2 θ + \sqrt{3} \tan θ \tan 2 θ = \sqrt{3}$

$\Rightarrow \tan θ + \tan 2 θ = \sqrt{3} - \sqrt{3} \tan θ \tan 2 θ$

$\Rightarrow \tan θ + \tan 2 θ = \sqrt{3} (1 - \tan θ \tan 2 θ)$

$\Rightarrow \frac{\tan θ + \tan 2 θ}{1 - \tan θ \tan 2 θ} = \sqrt{3}$

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

$\Rightarrow \tan (θ + 2 θ) = \sqrt{3}$

We know that $\tan \frac{π}{3} = \sqrt{3}$ . Therefore, we get

$\Rightarrow \tan (3 θ) = \tan \frac{π}{3}$

We know that if $\tan θ = \tan α$ then $θ = n π + α$ , $n \in Z$

$\Rightarrow 3 θ = n π + \frac{π}{3}$

$\Rightarrow θ = \frac{n π}{3} + \frac{π}{9}$

Thus, the given statement is true.

If $\tan (π \cos θ) = \cot (π \sin θ)$ , then $\cos (θ - \frac{π}{4}) = \pm \frac{1}{2 \sqrt{2}}$ .

Ans: Given, $\tan (π \cos θ) = \cot (π \sin θ)$

We know that $\tan (90^{\circ} - θ) = \cot θ$ . So, we can write above written equation as

$\Rightarrow \tan (π \cos θ) = \tan (\frac{π}{2} - π \sin θ)$

$\Rightarrow π \cos θ = \frac{π}{2} - π \sin θ$

$\Rightarrow π \cos θ + π \sin θ = \frac{π}{2}$

$\Rightarrow \cos θ + \sin θ = \frac{1}{2}$

Divide whole equation by $\sqrt{2}$ .

$\Rightarrow \frac{1}{\sqrt{2}} \cos θ + \frac{1}{\sqrt{2}} \sin θ = \frac{1}{2 \sqrt{2}}$

We know that $\sin \frac{π}{4} = \frac{1}{\sqrt{2}}$ and $\cos \frac{π}{4} = \frac{1}{\sqrt{2}}$ . Therefore, we can write above written equation as,

$\Rightarrow \cos \frac{π}{4} \cos θ + \sin \frac{π}{4} \sin θ = \frac{1}{2 \sqrt{2}}$

$\Rightarrow \cos θ \cos \frac{π}{4} + \sin θ \sin \frac{π}{4} = \frac{1}{2 \sqrt{2}}$

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

$\Rightarrow \cos (θ - \frac{π}{4}) = \pm \frac{1}{2 \sqrt{2}}$

Thus, the given statement is true.

In the following match each item given under the column $C_{1}$ to its correct answer given under the column $C_{2}$ .

$\sin (x + y) \sin (x - y)$ (i) $\cos^{2} x - \cos^{2} y$
$\cos (x + y) \cos (x - y)$ (ii) $\frac{1 - \tan θ}{1 + \tan θ}$
$\cot (\frac{π}{4} + θ)$ (iii) $\frac{1 - \tan θ}{1 + \tan θ}$
$\tan (\frac{π}{4} + θ)$ (iv) $\sin^{2} x - \sin^{2} y$

Ans:

$\sin (x + y) \sin (x - y)$

We know that $\sin (x + y) \sin (x - y) = \sin^{2} x - \sin^{2} y$

$∴ (a) \leftrightarrow (i v)$

$\cos (x + y) \cos (x - y)$

We know that $\cos (x + y) \cos (x - y) = \cos^{2} x - \cos^{2} y$

$∴ (b) \leftrightarrow (i)$

$\cot (\frac{π}{4} + θ)$

We know that $\cot (x + y) = \frac{\cot x . \cot y - 1}{\cot y + \cot x}$ . Therefore, we get

$\Rightarrow \cot (\frac{π}{4} + θ) = \frac{\cot \frac{π}{4} . \cot θ - 1}{\cot θ + \cot \frac{π}{4}}$

Substitute $\cot \frac{π}{4} = 1$

$\Rightarrow \cot (\frac{π}{4} + θ) = \frac{\cot θ - 1}{\cot θ + 1}$

Divide RHS by $\cot θ$

$\Rightarrow \cot (\frac{π}{4} + θ) = \frac{1 - \frac{1}{\cot θ}}{1 + \frac{1}{\cot θ}}$

$\Rightarrow \cot (\frac{π}{4} + θ) = \frac{1 - \tan θ}{1 + \tan θ}$

$∴ (c) \leftrightarrow (i i)$

$\tan (\frac{π}{4} + θ)$

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

$\Rightarrow \tan (\frac{π}{4} + θ) = \frac{\tan \frac{π}{4} + \tan θ}{1 - \tan \frac{π}{4} . \tan θ}$

Substitute $\tan \frac{π}{4} = 1$

$\Rightarrow \tan (\frac{π}{4} + θ) = \frac{1 + \tan θ}{1 - \tan θ}$

$∴ (d) \leftrightarrow (i i i)$

Class 11 Maths Chapter 3 Trigonometric Functions

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Concepts of Chapter 3

In this chapter, students will learn and solve exemplar problems based on topics as follows:

The relation between degree and radian
Trigonometric functions
Domain and range of trigonometric functions
Sine, cosine, and tangent of some angles less than 90°
Allied or related angles
Functions of negative angles
Some formulae regarding compound angles
Trigonometric equations
General Solution of Trigonometric Equations

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