NCERT Solutions for Class 10 Maths Chapter 8 Exercise 8.4

NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry (Ex 8.4) Exercise 8.4

NCERT Solutions for Class 10 Maths ex 8.4 designed by the subject matter experts at Vedantu provide a detailed explanation of every sum covered in the exercise. Our experts have prepared the solutions to each question in Class 10 Chapter 8 Exercise 8.4 in an easy step-by-step manner. By going through these Maths NCERT Solutions Class 10 Ex 8.4, you should be able to solve and analyze sums from trigonometric identities. There are a total of 5 questions in Maths Class 10 Chapter 8 Exercise 8.4 with many subparts, need long-winded derivations which are explained clearly by breaking down each problem in multiple steps. All the NCERT Book Solutions are prepared by experienced and highly qualified teaching staff. You can also find NCERT Solutions for Class 10 Science on Vedantu.

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Chapter 8 - Introduction to Trigonometry part-1

Access NCERT Solutions for Class 8 Maths Chapter 8 – Introduction to Trigonometry

Refer to Page 1 - 22 for Exercise 8.4

Exercise 8.4

1. Express the trigonometric ratios $\sin A,\sec A$ and $\tan A$ in terms of $\cot A$.

Ans: For a right triangle we have an identity ${{\operatorname{cosec}}^{2}}A=1+{{\cot }^{2}}A$.

Let us consider the above identity, we get

${{\operatorname{cosec}}^{2}}A=1+{{\cot }^{2}}A$

Now, reciprocating both sides we get

$\Rightarrow \dfrac{1}{{{\operatorname{cosec}}^{2}}A}=\dfrac{1}{1+{{\cot }^{2}}A}$

Now, we know that $\dfrac{1}{{{\operatorname{cosec}}^{2}}A}={{\sin }^{2}}A$, we get

$\Rightarrow {{\sin }^{2}}A=\dfrac{1}{1+{{\cot }^{2}}A}$

$\Rightarrow \sin A=\pm \dfrac{1}{\sqrt{1+{{\cot }^{2}}A}}$

Now, we know that sine value will be negative for angles greater than $180{}^\circ $, for a triangle sine value is always positive with respect to an angle. Then we will consider only positive value.

$\therefore \sin A=\dfrac{1}{\sqrt{1+{{\cot }^{2}}A}}$ 

We know that $\tan A=\dfrac{1}{\cot A}$ 

Also, we will use the identity ${{\sec }^{2}}A=1+{{\tan }^{2}}A$, we get

${{\sec }^{2}}A=1+{{\tan }^{2}}A$

$\Rightarrow {{\sec }^{2}}A=1+\dfrac{1}{{{\cot }^{2}}A}$

$\Rightarrow {{\sec }^{2}}A=\dfrac{{{\cot }^{2}}A+1}{{{\cot }^{2}}A}$

\[\Rightarrow \sec A=\dfrac{\sqrt{{{\cot }^{2}}A+1}}{\sqrt{{{\cot }^{2}}A}}\]

\[\therefore \sec A=\dfrac{\sqrt{{{\cot }^{2}}A+1}}{\cot A}\]


2. Write All the Other Trigonometric Ratios of $\angle A$ in terms of $\sec A$.

Ans: 

We know that $\cos A=\dfrac{1}{\sec A}$.

$\therefore \cos A=\dfrac{1}{\sec A}$

For a right triangle we have an identity ${{\sin }^{2}}A+{{\cos }^{2}}A=1$.

Let us consider the above identity, we get

${{\sin }^{2}}A+{{\cos }^{2}}A=1$

Now, we know that $\cos A=\dfrac{1}{\sec A}$, we get

$\Rightarrow {{\sin }^{2}}A=1-{{\cos }^{2}}A$

$\Rightarrow {{\sin }^{2}}A=1-\dfrac{1}{{{\sec }^{2}}A}$

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

$\therefore \sin A=\dfrac{\sqrt{{{\sec }^{2}}A-1}}{\sec A}$

Also, we will use the identity ${{\sec }^{2}}A=1+{{\tan }^{2}}A$, we get

${{\tan }^{2}}A={{\sec }^{2}}A-1$

$\therefore \tan A=\sqrt{{{\sec }^{2}}A-1}$ 

Now, we know that $\cot A=\dfrac{\cos A}{\sin A}$, we get

$\Rightarrow \cot A=\dfrac{\dfrac{1}{\sec A}}{\dfrac{\sqrt{{{\sec }^{2}}A-1}}{\sec A}}$

$\therefore \cot A=\dfrac{1}{\sqrt{{{\sec }^{2}}A-1}}$

We know that $cosecA=\dfrac{1}{\sin A}$, we get

 $\therefore cosecA=\dfrac{\sec A}{\sqrt{{{\sec }^{2}}A-1}}$


3. Evaluate the Following:

  1. $\dfrac{{{\sin }^{2}}63{}^\circ +{{\sin }^{2}}27{}^\circ }{{{\cos }^{2}}17{}^\circ +{{\cos }^{2}}73{}^\circ }$ 

Ans: The given expression is $\dfrac{{{\sin }^{2}}63{}^\circ +{{\sin }^{2}}27{}^\circ }{{{\cos }^{2}}17{}^\circ +{{\cos }^{2}}73{}^\circ }$.

The above expression can be written as 

$\dfrac{{{\sin }^{2}}63{}^\circ +{{\sin }^{2}}27{}^\circ }{{{\cos }^{2}}17{}^\circ +{{\cos }^{2}}73{}^\circ }=\dfrac{{{\left[ \sin \left( 90{}^\circ -27{}^\circ  \right) \right]}^{2}}+{{\sin }^{2}}27{}^\circ }{{{\left[ \cos \left( 90{}^\circ -73{}^\circ  \right) \right]}^{2}}+{{\cos }^{2}}73{}^\circ }$

Now, we can apply the identity $\cos \left( 90{}^\circ -\theta  \right)=\sin \theta $ and $\sin \left( 90{}^\circ -\theta  \right)=\cos \theta $, we get

$\Rightarrow \dfrac{{{\sin }^{2}}63{}^\circ +{{\sin }^{2}}27{}^\circ }{{{\cos }^{2}}17{}^\circ +{{\cos }^{2}}73{}^\circ }=\dfrac{{{\cos }^{2}}27{}^\circ +{{\sin }^{2}}27{}^\circ }{{{\sin }^{2}}73{}^\circ +{{\cos }^{2}}73{}^\circ }$

Now, by applying the trigonometric identity ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$, we get

$\Rightarrow \dfrac{{{\sin }^{2}}63{}^\circ +{{\sin }^{2}}27{}^\circ }{{{\cos }^{2}}17{}^\circ +{{\cos }^{2}}73{}^\circ }=\dfrac{1}{1}$ 

$\therefore \dfrac{{{\sin }^{2}}63{}^\circ +{{\sin }^{2}}27{}^\circ }{{{\cos }^{2}}17{}^\circ +{{\cos }^{2}}73{}^\circ }=1$  


  1. $\sin 25{}^\circ \cos 65{}^\circ +\cos 25{}^\circ \sin 65{}^\circ $ 

Ans: The given expression is $\sin 25{}^\circ \cos 65{}^\circ +\cos 25{}^\circ \sin 65{}^\circ $ .

The above expression can be written as 

$\sin 25{}^\circ \cos 65{}^\circ +\cos 25{}^\circ \sin 65{}^\circ =\sin 25{}^\circ \cos \left( 90{}^\circ -25{}^\circ  \right)+\cos 25{}^\circ \sin \left( 90{}^\circ -25{}^\circ  \right)$

Now, we can apply the identity $\cos \left( 90{}^\circ -\theta  \right)=\sin \theta $ and $\sin \left( 90{}^\circ -\theta  \right)=\cos \theta $, we get

\[\Rightarrow \sin 25{}^\circ \cos 65{}^\circ +\cos 25{}^\circ \sin 65{}^\circ =\sin 25{}^\circ \sin 25{}^\circ +\cos 25{}^\circ \cos 25{}^\circ \]

\[\Rightarrow \sin 25{}^\circ \cos 65{}^\circ +\cos 25{}^\circ \sin 65{}^\circ ={{\sin }^{2}}25{}^\circ +{{\cos }^{2}}25{}^\circ \]

Now, by applying the trigonometric identity ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$, we get

\[\Rightarrow \sin 25{}^\circ \cos 65{}^\circ +\cos 25{}^\circ \sin 65{}^\circ =1\] 

\[\therefore \sin 25{}^\circ \cos 65{}^\circ +\cos 25{}^\circ \sin 65{}^\circ =1\]

  

4. Choose the Correct Option and Justify Your Choice:

  1. \[9{{\sec }^{2}}A-9{{\tan }^{2}}A=\] …….

  1. $1$ 

  2. $9$

  3. $8$ 

  4. $0$

Ans: The given expression is $9{{\sec }^{2}}A-9{{\tan }^{2}}A$.

The given expression can be written as 

$\Rightarrow 9{{\sec }^{2}}A-9{{\tan }^{2}}A=9\left( {{\sec }^{2}}A-{{\tan }^{2}}A \right)$

Now, we will use the identity ${{\sec }^{2}}A=1+{{\tan }^{2}}A$, we get

${{\sec }^{2}}A-{{\tan }^{2}}A=1$

$\Rightarrow 9{{\sec }^{2}}A-9{{\tan }^{2}}A=9\left( 1 \right)$

$\therefore 9{{\sec }^{2}}A-9{{\tan }^{2}}A=9$

Therefore, option (B) is the correct answer.


  1. $\left( 1+\tan \theta +\sec \theta  \right)\left( 1+\cot \theta -\operatorname{cosec}\theta  \right)$ 

  1. $0$

  2. $1$

  3. $2$

  4. $-1$

Ans: The given expression is $\left( 1+\tan \theta +\sec \theta  \right)\left( 1+\cot \theta -\operatorname{cosec}\theta  \right)$.

We know that the trigonometric functions have values as:

$\tan \theta =\dfrac{\sin \theta }{\cos \theta }$, $\cot \theta =\dfrac{\cos \theta }{\sin \theta }=\dfrac{1}{\tan \theta }$, $cosec\theta =\dfrac{1}{\sin \theta }$ and $sec\theta =\dfrac{1}{\cos \theta }$

Substituting these values in the given expression, we get

$\Rightarrow \left( 1+\tan \theta +\sec \theta  \right)\left( 1+\cot \theta -\operatorname{cosec}\theta  \right)=\left( 1+\dfrac{\sin \theta }{\cos \theta }+\dfrac{1}{\cos \theta } \right)\left( 1+\dfrac{\cos \theta }{\sin \theta }-\dfrac{1}{\sin \theta } \right)$

$\Rightarrow \left( 1+\tan \theta +\sec \theta  \right)\left( 1+\cot \theta -\operatorname{cosec}\theta  \right)=\left( \dfrac{\cos \theta +\sin \theta +1}{\cos \theta } \right)\left( \dfrac{\sin \theta +\cos \theta -1}{\sin \theta } \right)$

Now, by applying the identity $\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$, we get

$\Rightarrow \left( 1+\tan \theta +\sec \theta  \right)\left( 1+\cot \theta -\operatorname{cosec}\theta  \right)=\dfrac{{{\left( \sin \theta +\cos \theta  \right)}^{2}}-{{1}^{2}}}{\sin \theta \cos \theta }$

$\Rightarrow \left( 1+\tan \theta +\sec \theta  \right)\left( 1+\cot \theta -\operatorname{cosec}\theta  \right)=\dfrac{{{\sin }^{2}}\theta +{{\cos }^{2}}\theta +2\sin \theta \cos \theta -1}{\sin \theta \cos \theta }$

Now, by applying the trigonometric identity ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$, we get

$\Rightarrow \left( 1+\tan \theta +\sec \theta  \right)\left( 1+\cot \theta -\operatorname{cosec}\theta  \right)=\dfrac{1+2\sin \theta \cos \theta -1}{\sin \theta \cos \theta }$

$\Rightarrow \left( 1+\tan \theta +\sec \theta  \right)\left( 1+\cot \theta -\operatorname{cosec}\theta  \right)=\dfrac{2\sin \theta \cos \theta }{\sin \theta \cos \theta }$

$\therefore \left( 1+\tan \theta +\sec \theta  \right)\left( 1+\cot \theta -\operatorname{cosec}\theta  \right)=2$

Therefore, option (C) is the correct answer.


  1. $\left( \sec A+\tan A \right)\left( 1-\sin A \right)=$ ………

  1. $\sec A$ 

  2. $\sin A$ 

  3. $cosecA$ 

  4. $\cos A$ 

Ans: Given expression is $\left( \sec A+\tan A \right)\left( 1-\sin A \right)$.

We know that $\tan \theta =\dfrac{\sin \theta }{\cos \theta }$ and $sec\theta =\dfrac{1}{\cos \theta }$

Substituting these values in the given expression, we get

$\left( \sec A+\tan A \right)\left( 1-\sin A \right)=\left( \dfrac{1}{\cos A}+\dfrac{\sin A}{\cos A} \right)\left( 1-\sin A \right)$

$\Rightarrow \left( \sec A+\tan A \right)\left( 1-\sin A \right)=\left( \dfrac{1+\sin A}{\cos A} \right)\left( 1-\sin A \right)$

$\Rightarrow \left( \sec A+\tan A \right)\left( 1-\sin A \right)=\left( \dfrac{\left( 1+\sin A \right)\left( 1-\sin A \right)}{\cos A} \right)$

Now, by applying the identity $\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$, we get

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

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

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

$\therefore \left( \sec A+\tan A \right)\left( 1-\sin A \right)=\cos A$

Therefore, option (D) is the correct answer.


  1. $\dfrac{1+{{\tan }^{2}}A}{1+{{\cot }^{2}}A}$ 

  1. ${{\sec }^{2}}A$ 

  2. $-1$ 

  3. ${{\cot }^{2}}A$ 

  4. ${{\tan }^{2}}A$ 

Ans: Given expression is $\dfrac{1+{{\tan }^{2}}A}{1+{{\cot }^{2}}A}$.

We know that the trigonometric functions have values as:

$\tan \theta =\dfrac{\sin \theta }{\cos \theta }$ and $\cot \theta =\dfrac{\cos \theta }{\sin \theta }=\dfrac{1}{\tan \theta }$.

Substituting these values in the given expression, we get

$\dfrac{1+{{\tan }^{2}}A}{1+{{\cot }^{2}}A}=\dfrac{1+\dfrac{{{\sin }^{2}}A}{{{\cos }^{2}}A}}{1+\dfrac{{{\cos }^{2}}A}{{{\sin }^{2}}A}}$

$\Rightarrow \dfrac{1+{{\tan }^{2}}A}{1+{{\cot }^{2}}A}=\dfrac{\dfrac{{{\cos }^{2}}A+{{\sin }^{2}}A}{{{\cos }^{2}}A}}{\dfrac{{{\sin }^{2}}A+{{\cos }^{2}}A}{{{\sin }^{2}}A}}$

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

$\Rightarrow \dfrac{1+{{\tan }^{2}}A}{1+{{\cot }^{2}}A}=\dfrac{\dfrac{1}{{{\cos }^{2}}A}}{\dfrac{1}{{{\sin }^{2}}A}}$

$\Rightarrow \dfrac{1+{{\tan }^{2}}A}{1+{{\cot }^{2}}A}=\dfrac{{{\sin }^{2}}A}{{{\cos }^{2}}A}$

$\Rightarrow \dfrac{1+{{\tan }^{2}}A}{1+{{\cot }^{2}}A}={{\tan }^{2}}A$

Therefore, option (D) is the correct answer.


5. Prove the Following Identities, Where the Angles Involved are Acute Angles for Which the Expressions are Defined.

  1. ${{\left( cosec\theta -cot\theta  \right)}^{2}}=\dfrac{1-\cos \theta }{1+\cos \theta }$ 

Ans: Given expression is ${{\left( cosec\theta -\cot \theta  \right)}^{2}}=\dfrac{1-\cos \theta }{1+\cos \theta }$.

Let us consider the LHS of the given expression, we get

$LHS={{\left( cosec\theta -\cot \theta  \right)}^{2}}$

Now, we know that $\cot \theta =\dfrac{\cos \theta }{\sin \theta }=\dfrac{1}{\tan \theta }$ and $cosec\theta =\dfrac{1}{\sin \theta }$.

By substituting the values, we get

$\Rightarrow {{\left( cosec\theta -\cot \theta  \right)}^{2}}={{\left( \dfrac{1}{\sin \theta }-\dfrac{\cos \theta }{\sin \theta } \right)}^{2}}$

$\Rightarrow {{\left( cosec\theta -\cot \theta  \right)}^{2}}={{\left( \dfrac{1-\cos \theta }{\sin \theta } \right)}^{2}}$

$\Rightarrow {{\left( cosec\theta -\cot \theta  \right)}^{2}}=\dfrac{{{\left( 1-\cos \theta  \right)}^{2}}}{{{\sin }^{2}}\theta }$

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

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

Now, by applying the identity $\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$, we get

$\Rightarrow {{\left( cosec\theta -\cot \theta  \right)}^{2}}=\dfrac{{{\left( 1-\cos \theta  \right)}^{2}}}{\left( 1-\cos \theta  \right)\left( 1+\cos \theta  \right)}$

$\Rightarrow {{\left( cosec\theta -\cot \theta  \right)}^{2}}=\dfrac{\left( 1-\cos \theta  \right)}{\left( 1+\cos \theta  \right)}$

$\Rightarrow {{\left( cosec\theta -\cot \theta  \right)}^{2}}=RHS$

$\therefore {{\left( cosec\theta -\cot \theta  \right)}^{2}}=\dfrac{1-\cos \theta }{1+\cos \theta }$

Hence proved


  1. $\dfrac{\cos A}{1+\sin A}+\dfrac{1+\sin A}{\cos A}=2\sec A$

Ans: Given expression is $\dfrac{\cos A}{1+\sin A}+\dfrac{1+\sin A}{\cos A}=2\sec A$.

Let us consider the LHS of the given expression, we get

$LHS=\dfrac{\cos A}{1+\sin A}+\dfrac{1+\sin A}{\cos A}$

Now, taking LCM, we get

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

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

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

$\Rightarrow \dfrac{\cos A}{1+\sin A}+\dfrac{1+\sin A}{\cos A}=\dfrac{1+2\sin A+1}{\left( 1+\sin A \right)\cos A}$

$\Rightarrow \dfrac{\cos A}{1+\sin A}+\dfrac{1+\sin A}{\cos A}=\dfrac{2+2\sin A}{\left( 1+\sin A \right)\cos A}$

$\Rightarrow \dfrac{\cos A}{1+\sin A}+\dfrac{1+\sin A}{\cos A}=\dfrac{2\left( 1+\sin A \right)}{\left( 1+\sin A \right)\cos A}$

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

We know that $sec\theta =\dfrac{1}{\cos \theta }$, we get

$\Rightarrow \dfrac{\cos A}{1+\sin A}+\dfrac{1+\sin A}{\cos A}=2\sec A$

\[\Rightarrow \dfrac{\cos A}{1+\sin A}+\dfrac{1+\sin A}{\cos A}=RHS\]

\[\therefore \dfrac{\cos A}{1+\sin A}+\dfrac{1+\sin A}{\cos A}=2\sec A\]

Hence proved


  1. $\dfrac{\tan \theta }{1-\cot \theta }+\dfrac{\cot \theta }{1-\tan \theta }=1+\sec \theta cosec\theta $ 

Ans: Given expression is $\dfrac{\tan \theta }{1-\cot \theta }+\dfrac{\cot \theta }{1-\tan \theta }=1+\sec \theta \operatorname{cosec}\theta $.

Let us consider the LHS of the given expression, we get

$LHS=\dfrac{\tan \theta }{1-\cot \theta }+\dfrac{\cot \theta }{1-\tan \theta }$

Now, we know that $\tan \theta =\dfrac{\sin \theta }{\cos \theta }$ and $\cot \theta =\dfrac{\cos \theta }{\sin \theta }=\dfrac{1}{\tan \theta }$.

By substituting the values, we get

\[\Rightarrow \dfrac{\tan \theta }{1-\cot \theta }+\dfrac{\cot \theta }{1-\tan \theta }=\left( \dfrac{\dfrac{\sin \theta }{\cos \theta }}{1-\dfrac{\cos \theta }{\sin \theta }}+\dfrac{\dfrac{\cos \theta }{\sin \theta }}{1-\dfrac{\sin \theta }{\cos \theta }} \right)\]

\[\Rightarrow \dfrac{\tan \theta }{1-\cot \theta }+\dfrac{\cot \theta }{1-\tan \theta }=\left( \dfrac{\dfrac{\sin \theta }{\cos \theta }}{\dfrac{\sin \theta -\cos \theta }{\sin \theta }}+\dfrac{\dfrac{\cos \theta }{\sin \theta }}{\dfrac{\cos \theta -\sin \theta }{\cos \theta }} \right)\]

 \[\Rightarrow \dfrac{\tan \theta }{1-\cot \theta }+\dfrac{\cot \theta }{1-\tan \theta }=\left( \dfrac{{{\sin }^{2}}\theta }{\cos \theta \left( \sin \theta -\cos \theta  \right)}+\dfrac{{{\cos }^{2}}\theta }{\sin \theta \left( \sin \theta -\cos \theta  \right)} \right)\]

\[\Rightarrow \dfrac{\tan \theta }{1-\cot \theta }+\dfrac{\cot \theta }{1-\tan \theta }=\dfrac{1}{\left( \sin \theta -\cos \theta  \right)}\left( \dfrac{{{\sin }^{2}}\theta }{\cos \theta }+\dfrac{{{\cos }^{2}}\theta }{\sin \theta } \right)\]

\[\Rightarrow \dfrac{\tan \theta }{1-\cot \theta }+\dfrac{\cot \theta }{1-\tan \theta }=\dfrac{1}{\left( \sin \theta -\cos \theta  \right)}\left( \dfrac{{{\sin }^{3}}\theta -{{\cos }^{3}}\theta }{\sin \theta \cos \theta } \right)\]

Now, by applying the identity \[{{a}^{3}}-{{b}^{3}}=\left( a-b \right)\left( {{a}^{2}}+ab+{{b}^{2}} \right)\], we get

\[\Rightarrow \dfrac{\tan \theta }{1-\cot \theta }+\dfrac{\cot \theta }{1-\tan \theta }=\dfrac{1}{\left( \sin \theta -\cos \theta  \right)}\left[ \dfrac{\left( \sin \theta -\cos \theta  \right)\left( {{\sin }^{2}}\theta +{{\cos }^{2}}\theta +\sin \theta \cos \theta  \right)}{\sin \theta \cos \theta } \right]\]

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

\[\Rightarrow \dfrac{\tan \theta }{1-\cot \theta }+\dfrac{\cot \theta }{1-\tan \theta }=\dfrac{1}{\left( \sin \theta -\cos \theta  \right)}\left[ \dfrac{\left( \sin \theta -\cos \theta  \right)\left( 1+\sin \theta \cos \theta  \right)}{\sin \theta \cos \theta } \right]\]

\[\Rightarrow \dfrac{\tan \theta }{1-\cot \theta }+\dfrac{\cot \theta }{1-\tan \theta }=\dfrac{\left( 1+\sin \theta \cos \theta  \right)}{\sin \theta \cos \theta }\]

\[\Rightarrow \dfrac{\tan \theta }{1-\cot \theta }+\dfrac{\cot \theta }{1-\tan \theta }=\dfrac{1}{\sin \theta \cos \theta }+\dfrac{\sin \theta \cos \theta }{\sin \theta \cos \theta }\]

\[\Rightarrow \dfrac{\tan \theta }{1-\cot \theta }+\dfrac{\cot \theta }{1-\tan \theta }=\dfrac{1}{\sin \theta \cos \theta }+1\]

We know that $cosec\theta =\dfrac{1}{\sin \theta }$ and $sec\theta =\dfrac{1}{\cos \theta }$, we get

\[\Rightarrow \dfrac{\tan \theta }{1-\cot \theta }+\dfrac{\cot \theta }{1-\tan \theta }=\sec \theta cosec\theta +1\]

\[\Rightarrow \dfrac{\tan \theta }{1-\cot \theta }+\dfrac{\cot \theta }{1-\tan \theta }=1+\sec \theta cosec\theta \]

\[\Rightarrow \dfrac{\tan \theta }{1-\cot \theta }+\dfrac{\cot \theta }{1-\tan \theta }=RHS\]

\[\therefore \dfrac{\tan \theta }{1-\cot \theta }+\dfrac{\cot \theta }{1-\tan \theta }=1+\sec \theta cosec\theta \]

Hence proved


  1. $\dfrac{1+\sec A}{\sec A}=\dfrac{{{\sin }^{2}}A}{1-\cos A}$ 

Ans: Given expression is $\dfrac{1+\sec A}{\sec A}=\dfrac{{{\sin }^{2}}A}{1-\cos A}$.

Let us consider the LHS of the given expression, we get

$LHS=\dfrac{1+\sec A}{\sec A}$

Now, we know that $\sec \theta =\dfrac{1}{\cos \theta }$.

By substituting the value, we get

\[\Rightarrow \dfrac{1+\sec A}{\sec A}=\dfrac{1+\dfrac{1}{\cos A}}{\dfrac{1}{\cos A}}\]

\[\Rightarrow \dfrac{1+\sec A}{\sec A}=\dfrac{\dfrac{\cos A+1}{\cos A}}{\dfrac{1}{\cos A}}\]

 \[\Rightarrow \dfrac{1+\sec A}{\sec A}=\cos A+1\]

Multiply and divide by $\left( 1-\cos A \right)$, we get

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

Now, by applying the identity $\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$, we get

\[\Rightarrow \dfrac{1+\sec A}{\sec A}=\dfrac{1-{{\cos }^{2}}A}{\left( 1-\cos A \right)}\]

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

\[\Rightarrow \dfrac{1+\sec A}{\sec A}=\dfrac{{{\sin }^{2}}A}{\left( 1-\cos A \right)}\]

\[\Rightarrow \dfrac{1+\sec A}{\sec A}=RHS\]

\[\therefore \dfrac{1+\sec A}{\sec A}=\dfrac{{{\sin }^{2}}A}{1-\cos A}\]

Hence proved


  1. $\dfrac{\cos A-\sin A+1}{\cos A+\sin A-1}=cosecA+\cot A$ 

Ans: Given expression is $\dfrac{\cos A-\sin A+1}{\cos A+\sin A-1}=cosecA+\cot A$.

Now, let us consider the LHS of the given expression, we get

$LHS=\dfrac{\cos A-\sin A+1}{\cos A+\sin A-1}$

Dividing numerator and denominator by $\sin A$, we get

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

Now, we know that $\cot \theta =\dfrac{\cos \theta }{\sin \theta }=\dfrac{1}{\tan \theta }$ and $cosec\theta =\dfrac{1}{\sin \theta }$, we get

$\Rightarrow \dfrac{\cos A-\sin A+1}{\cos A+\sin A-1}=\dfrac{\cot A-1+\operatorname{cosec}A}{\cot A+1-\operatorname{cosec}A}$

Now, by applying the identity ${{\operatorname{cosec}}^{2}}A=1+{{\cot }^{2}}A$, substitute $1={{\cot }^{2}}A-{{\operatorname{cosec}}^{2}}A$, we get

$\Rightarrow \dfrac{\cos A-\sin A+1}{\cos A+\sin A-1}=\dfrac{\cot A-\left( {{\cot }^{2}}A-{{\operatorname{cosec}}^{2}}A \right)+\operatorname{cosec}A}{\cot A+{{\cot }^{2}}A-{{\operatorname{cosec}}^{2}}A-\operatorname{cosec}A}$

$\Rightarrow \dfrac{\cos A-\sin A+1}{\cos A+\sin A-1}=\dfrac{\cot A-{{\cot }^{2}}A+{{\operatorname{cosec}}^{2}}A+\operatorname{cosec}A}{\cot A+{{\cot }^{2}}A-{{\operatorname{cosec}}^{2}}A-\operatorname{cosec}A}$

$\Rightarrow \dfrac{\cos A-\sin A+1}{\cos A+\sin A-1}=\dfrac{{{\left( \cot A-1+\operatorname{cosec}A \right)}^{2}}}{{{\cot }^{2}}A-1+{{\operatorname{cosec}}^{2}}A+2\operatorname{cosec}A}$

$\Rightarrow \dfrac{\cos A-\sin A+1}{\cos A+\sin A-1}=\dfrac{{{\left( \cot A-1+\operatorname{cosec}A \right)}^{2}}}{{{\cot }^{2}}A-1+{{\operatorname{cosec}}^{2}}A+2\operatorname{cosec}A}$

$\Rightarrow \dfrac{\cos A-\sin A+1}{\cos A+\sin A-1}=\dfrac{2{{\operatorname{cosec}}^{2}}A+2\cot A\operatorname{cosec}A-2\cot A-2\operatorname{cosec}A}{{{\cot }^{2}}A-1+{{\operatorname{cosec}}^{2}}A+2\operatorname{cosec}A}$

$\Rightarrow \dfrac{\cos A-\sin A+1}{\cos A+\sin A-1}=\dfrac{2\operatorname{cosec}A\left( \cot A-\operatorname{cosec}A \right)-2\left( \cot A-\operatorname{cosec}A \right)}{{{\cot }^{2}}A-1+{{\operatorname{cosec}}^{2}}A+2\operatorname{cosec}A}$

$\Rightarrow \dfrac{\cos A-\sin A+1}{\cos A+\sin A-1}=\dfrac{\left( 2\operatorname{cosec}A-2 \right)\left( \cot A-\operatorname{cosec}A \right)}{1-1+2\operatorname{cosec}A}$

$\Rightarrow \dfrac{\cos A-\sin A+1}{\cos A+\sin A-1}=\dfrac{\left( 2\operatorname{cosec}A-2 \right)\left( \cot A-\operatorname{cosec}A \right)}{2\operatorname{cosec}A}$

$\Rightarrow \dfrac{\cos A-\sin A+1}{\cos A+\sin A-1}=\operatorname{cosec}A+\cot A$

$\Rightarrow \dfrac{\cos A-\sin A+1}{\cos A+\sin A-1}=RHS$

$\therefore \dfrac{\cos A-\sin A+1}{\cos A+\sin A-1}=\operatorname{cosec}A+\cot A$ 

Hence proved


  1. $\sqrt{\dfrac{1+\sin A}{1-\sin A}}=\sec A+\tan A$ 

Ans: Given expression is $\sqrt{\dfrac{1+\sin A}{1-\sin A}}=\sec A+\tan A$.

Let us consider the LHS of the given expression, we get

$LHS=\sqrt{\dfrac{1+\sin A}{1-\sin A}}$

Now, multiply and divide the expression by $\sqrt{1+\sin A}$, we get

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

Now, by applying the identity $\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$, we get

\[\Rightarrow \sqrt{\dfrac{1+\sin A}{1-\sin A}}=\sqrt{\dfrac{{{\left( 1+\sin A \right)}^{2}}}{1-{{\sin }^{2}}A}}\]

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

\[\Rightarrow \sqrt{\dfrac{1+\sin A}{1-\sin A}}=\dfrac{1+\sin A}{\sqrt{{{\cos }^{2}}A}}\]

\[\Rightarrow \sqrt{\dfrac{1+\sin A}{1-\sin A}}=\dfrac{1+\sin A}{\cos A}\]

\[\Rightarrow \sqrt{\dfrac{1+\sin A}{1-\sin A}}=\dfrac{1}{\cos A}+\dfrac{\sin A}{\cos A}\]

\[\Rightarrow \sqrt{\dfrac{1+\sin A}{1-\sin A}}=\sec A+\tan A\]

\[\Rightarrow \sqrt{\dfrac{1+\sin A}{1-\sin A}}=RHS\]

$\therefore \sqrt{\dfrac{1+\sin A}{1-\sin A}}=\sec A+\tan A$ 

Hence proved


  1. $\dfrac{\sin \theta -2{{\sin }^{3}}\theta }{2\cos \theta -\cos \theta }=\tan \theta $ 

Ans: Given expression is $\dfrac{\sin \theta -2{{\sin }^{3}}\theta }{2\cos \theta -\cos \theta }=\tan \theta $.

Let us consider the LHS of the given expression, we get

$LHS=\dfrac{\sin \theta -2{{\sin }^{3}}\theta }{2\cos \theta -\cos \theta }$

Taking common terms out, we get

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

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

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

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

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

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

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

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

$\therefore \dfrac{\sin \theta -2{{\sin }^{3}}\theta }{2\cos \theta -\cos \theta }=\tan \theta $

Hence proved


  1. ${{\left( \sin A+cosecA \right)}^{2}}+{{\left( \cos A+secA \right)}^{2}}=7+{{\tan }^{2}}A+{{\cot }^{2}}A$ 

Ans: Given expression is ${{\left( \sin A+cosecA \right)}^{2}}+{{\left( \cos A+\sec A \right)}^{2}}=7+{{\tan }^{2}}A+{{\cot }^{2}}A$.

Let us consider the LHS of the given expression, we get

$LHS={{\left( \sin A+cosecA \right)}^{2}}+{{\left( \cos A+\sec A \right)}^{2}}$

Now, by applying the identity \[{{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}\], we get

\[\Rightarrow {{\left( \sin A+cosecA \right)}^{2}}+{{\left( \cos A+\sec A \right)}^{2}}={{\sin }^{2}}A+cosec{{A}^{2}}+2\sin AcosecA+{{\cos }^{2}}A+{{\sec }^{2}}A+2\cos A\sec A\]\[\Rightarrow {{\left( \sin A+cosecA \right)}^{2}}+{{\left( \cos A+\sec A \right)}^{2}}={{\sin }^{2}}A+{{\cos }^{2}}A+cosec{{A}^{2}}+{{\sec }^{2}}A+2\sin AcosecA+2\cos A\sec A\]We know that ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$, $cosec\theta =\dfrac{1}{\sin \theta }$ and $sec\theta =\dfrac{1}{\cos \theta }$, we get

\[\Rightarrow {{\left( \sin A+cosecA \right)}^{2}}+{{\left( \cos A+\sec A \right)}^{2}}=1+cose{{c}^{2}}\theta +{{\sec }^{2}}\theta +2\sin A\dfrac{1}{\sin A}+2\cos A\dfrac{1}{\cos A}\] 

\[\Rightarrow {{\left( \sin A+cosecA \right)}^{2}}+{{\left( \cos A+\sec A \right)}^{2}}=1+\left( 1+{{\cot }^{2}}A+1+{{\tan }^{2}}A \right)+2+2\]

\[\Rightarrow {{\left( \sin A+cosecA \right)}^{2}}+{{\left( \cos A+\sec A \right)}^{2}}=7+{{\tan }^{2}}A+{{\cot }^{2}}A\]

\[\Rightarrow {{\left( \sin A+cosecA \right)}^{2}}+{{\left( \cos A+\sec A \right)}^{2}}=RHS\]

\[\therefore {{\left( \sin A+cosecA \right)}^{2}}+{{\left( \cos A+\sec A \right)}^{2}}=7+{{\tan }^{2}}A+{{\cot }^{2}}A\]

Hence proved


  1. $\left( cosecA-\sin A \right)\left( \sec A-\cos A \right)=\dfrac{1}{\tan A+\cot A}$

Ans: Given expression is $\left( cosecA-\sin A \right)\left( \sec A-\cos A \right)=\dfrac{1}{\tan A+\cot A}$.

Let us consider the LHS of the given expression, we get

$LHS=\left( cosecA-\sin A \right)\left( \sec A-\cos A \right)$

We know that $cosec\theta =\dfrac{1}{\sin \theta }$ and $sec\theta =\dfrac{1}{\cos \theta }$, we get

$\Rightarrow \left( cosecA-\sin A \right)\left( \sec A-\cos A \right)=\left( \dfrac{1}{\sin A}-\sin A \right)\left( \dfrac{1}{\cos A}-\cos A \right)$

$\Rightarrow \left( cosecA-\sin A \right)\left( \sec A-\cos A \right)=\left( \dfrac{1-{{\sin }^{2}}A}{\sin A} \right)\left( \dfrac{1-{{\cos }^{2}}A}{\cos A} \right)$

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

$\Rightarrow \left( cosecA-\sin A \right)\left( \sec A-\cos A \right)=\left( \dfrac{{{\cos }^{2}}A}{\sin A} \right)\left( \dfrac{{{\sin }^{2}}A}{\cos A} \right)$

$\Rightarrow \left( cosecA-\sin A \right)\left( \sec A-\cos A \right)=\sin A\cos A$

Now, consider the RHS of the given expression, we get

$RHS=\dfrac{1}{\tan A+\cot A}$

Now, we know that $\tan \theta =\dfrac{\sin \theta }{\cos \theta }$ and $\cot \theta =\dfrac{\cos \theta }{\sin \theta }=\dfrac{1}{\tan \theta }$.

$\Rightarrow \dfrac{1}{\tan A+\cot A}=\dfrac{1}{\dfrac{\sin A}{\cos A}+\dfrac{\cos A}{\sin A}}$

$\Rightarrow \dfrac{1}{\tan A+\cot A}=\dfrac{1}{\dfrac{{{\sin }^{2}}A+{{\cos }^{2}}A}{\sin A\cos A}}$

$\Rightarrow \dfrac{1}{\tan A+\cot A}=\dfrac{\sin A\cos A}{{{\sin }^{2}}A+{{\cos }^{2}}A}$

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

$\Rightarrow \dfrac{1}{\tan A+\cot A}=\sin A\cos A$

Here, we get LHS=RHS

$\therefore \left( cosecA-\sin A \right)\left( \sec A-\cos A \right)=\dfrac{1}{\tan A+\cot A}$ 

Hence proved


  1. $\left( \dfrac{1+{{\tan }^{2}}A}{1+{{\cot }^{2}}A} \right)={{\left( \dfrac{1-\tan A}{1-\cot A} \right)}^{2}}$

Ans: Given expression is $\dfrac{1+{{\tan }^{2}}A}{1+{{\cot }^{2}}A}={{\left( \dfrac{1-\tan A}{1-\cot A} \right)}^{2}}$.

Let us consider the LHS of the given expression, we get

$LHS=\dfrac{1+{{\tan }^{2}}A}{1+{{\cot }^{2}}A}$

By applying the identities ${{\sec }^{2}}A=1+{{\tan }^{2}}A$ and ${{\operatorname{cosec}}^{2}}A=1+{{\cot }^{2}}A$, we get

$\Rightarrow \dfrac{1+{{\tan }^{2}}A}{1+{{\cot }^{2}}A}=\dfrac{se{{c}^{2}}A}{{{\operatorname{cosec}}^{2}}A}$

We know that $cosec\theta =\dfrac{1}{\sin \theta }$ and $sec\theta =\dfrac{1}{\cos \theta }$, we get

$\Rightarrow \dfrac{1+{{\tan }^{2}}A}{1+{{\cot }^{2}}A}=\dfrac{\dfrac{1}{{{\cos }^{2}}A}}{\dfrac{1}{{{\sin }^{2}}A}}$

$\Rightarrow \dfrac{1+{{\tan }^{2}}A}{1+{{\cot }^{2}}A}=\dfrac{{{\sin }^{2}}A}{{{\cos }^{2}}A}$

$\Rightarrow \dfrac{1+{{\tan }^{2}}A}{1+{{\cot }^{2}}A}={{\tan }^{2}}A$

Now, consider the RHS of the given expression, we get

$RHS={{\left( \dfrac{1-\tan A}{1-\cot A} \right)}^{2}}$

Now, we know that $\cot \theta =\dfrac{1}{\tan \theta }$, we get

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

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

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

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

Here, we get LHS=RHS

$\therefore \dfrac{1+{{\tan }^{2}}A}{1+{{\cot }^{2}}A}={{\left( \dfrac{1-\tan A}{1-\cot A} \right)}^{2}}$ 

Hence proved


CBSE Class 10 Maths Chapter 8 Exercise 8.4 Solutions Free PDF

The NCERT Solutions Class 10 Maths Chapter 8 Exercise 8.4 is now available in PDF format at the official website of Vedantu. The CBSE Class 10 Maths Exercise 8.4 Solutions PDF can be downloaded on your devices or printed out. Once you have it in this form, you can do a quick revision of all crucial formulas without needing any internet connectivity. You could also do a group study with your friends by using the hard copy of the solutions.

 

Class 10 Maths Exercise 8.4 Solutions

Exercise 8.4 Class 10 Question 1

In the first question of Class 10 Maths ex 8.4, you need to express the trigonometric ratios Sin, Sec, and Tan in terms of Cot of an angle. To solve this question, you would make use of trigonometric identities cos2θ = 1 + cot2θ and tan θ = sin θ/cos θ.


Exercise 8.4 Class 10 Question 2

The second question of this exercise is a simple one where you need to write down all the trigonometric ratios of an angle θ in terms of Sec θ. You will start from the trigonometric function Cos θ = 1/Secθ along with the trigonometric ratio Sin2θ + Cos2θ = 1 and transpose to derive the solution.


Exercise 8.4 Class 10 Question 3

This question of Class 10 Maths Trigonometry ex 8.4 has two parts where students need to find out the value of a few complex trigonometric functions involving different angles. By using various concepts learned in trigonometries like the trigonometric identity Sin2θ + Cos2θ = 1 and formulas like Sin (90൦ - θ) = Cos θ and Cos (90൦ - θ) = Sin θ, students can find the values of these two trigonometric equations.


Exercise 8.4 Class 10 Question 4

The 4th question of Trigonometry Class 10 Exercise 8.4 has 4 subparts which are multiple-choice questions. In this, a few trigonometric equations are given whose values have to be selected from the 4 options provided. Students need to choose the correct choice as well as provide the reasoning for choosing that particular answer. The Maths 8.4 Class 10 solution provided by our experts for these questions makes use of many trigonometric identities like:

  • tan2θ = 1 + sec 2θ

  • tan θ = sinθ/cosθ

  • Cotθ = 1/tanθ

  • Secθ = 1/Cosθ

  • Cosecθ = 1/Sinθ

  • Sin2 θ + Cos2 θ = 1

The algebraic equation a2 - b2 = (a + b) * (a - b) is also used in solving these equations.


Exercise 8.4 Class 10 Question 5

In Class 10 Maths Chapter 8 Exercise 8.4 Question 5 students are given 10 trigonometric identities to prove. The angles for each of those 10 identities are acute angles, i.e. less than 90o. Students would need to apply the following trigonometric relationships to solve the 10 trigonometric identities:

  • Cot θ = 1/tan θ = Sin θ/Cos θ 

  • Cosec θ = 1/Sin θ

  • Sin2 θ + Cos2 θ = 1

  • Sec θ = 1/ Cosθ

  • Cosec 2θ = 1 + Cot2 θ

  • tan θ = sin θ/cosθ

You would also use your knowledge of solving algebra by using equations like 

  • a2 - b2 = (a + b) * (a - b)  and

  • (a+ b)2 = a2 + 2ab + b2


Key Features of NCERT Solutions for Class 10 Maths Chapter 8 Exercise 8.4

The expert team of Vedantu has done an extensive research to give you solutions for Exercise 8 4 Class 10. These NCERT solutions Class 10 Maths 8.4 will be a great online resource for your maths preparation due to many reasons like:

  • The solution for all the questions of Class 10 8.4 is explained in detail by giving supporting formulas and theories. You can easily brush up on all the concepts quickly with our solution.

  • The PDF form of the solution available at the official website of Vedantu can be downloaded for offline access. It is also printable so that you have many modes of revising all the questions in this exercise quickly.

  • The answers are formulated by following the updated CBSE curriculum; hence going through the solution will surely secure high marks for you in your maths exams.

  • The team of Vedantu does not only provide solutions but is also available for consultation when you get stuck in between. They are prompt in answering your query so that you do not lose your precious exam time.

FAQs (Frequently Asked Questions)

Q1. How Would you Define Trigonometric Identities?

Ans: If an equation is true for all values of the variables involved in the equation, it is called an identity equation. Trigonometric identities are equations that involve trigonometric ratios of an angle which is true for all values of the angle or angles involved in the equation.

Q2. In a Right Angle Triangle ABC Where Angle B is 90 Degrees, how will you Prove that for any of the Acute Angles, say θ in that Triangle: Sin2θ + Cos2θ = 1.

Ans: If B is 90 degrees then the side AC is the hypotenuse of the right-angle triangle. So as per Pythagoras theorem:

AC2 = AB2 + BC2

If we divide each term in the above equation by AC2 we get:

AC2/AC2 = AB2/AC2 + BC2/AC2

I.e (AC/AC)2 = (AB/AC)2 + (BC/AC)2

So, 1 = Cos2θ + Sin2θ, where θ is the angle opposite to side BC.

Q3. How many difficult illustrations are there in 10th Maths Exercise 8.4?

Ans: In Class 10 Maths Exercise 8.4, you will be able to analyse and solve Trigonometry questions. There are in total five questions with different sub-parts, which need to be solved by different equations of Trigonometry. Different identities that need to be used in this exercise are:

  • Tan2θ = -1 + Sec2 θ

  • Tan θ = Sinθ/Cosθ

  • Cotθ = 1/tanθ

  • Secθ = 1/Cosθ

  • Cosecθ = 1/Sinθ

  • Sin2 θ + Cos2 θ = 1

Q4. Which is the best ever example in Exercise 8.4 of 10th Maths?

Ans: The best example from Class 10 Maths Exercise 8.4 is the question to prove 10 different trigonometric identities. As it is important to remember all the trigonometric identities, you can solve this question to revise all the important basic identities that will be used to solve the questions. You can make use of the basic identities of trigonometry and algebra to solve and prove the given identities.

Q5. Why is the chapter on Trigonometry considered to be the most important chapter?

Ans: The chapter on Trigonometry is of utmost importance in Class 10 Maths. You will get a large number of questions from this chapter in your exam. This chapter carries high weightage. If prepared well, this chapter can help you score several marks in the exam. If you practice enough questions with a deep understanding of the concept, it will definitely be a game-changer for your grades in Class 10.

Q6. Why do students choose NCERT Maths Class 10 Chapter 8 Exercise 8.4 Solutions?

Ans: Trigonometry can be a bit tricky because you have to illustrate, apply identities, and find the solutions by solving them. If you get stuck somewhere, you need a guide to understand what needs to be done next. Hence, students must solve all the questions themselves. NCERT Maths Class 10 Chapter 8 Exercise 8.4 Solutions are prepared by the subject experts with high accuracy. You can rely on them for the best tricks to solve different questions. You can find these solutions free of cost on the Vedantu website (vedantu.com) and also on our Vedantu Mobile app.

Q7. Where can I find the downloadable solutions for NCERT Class 10 Maths Exercise 8.4?

Ans: To find the downloadable solutions for NCERT Class 10 Maths Exercise 8.4, follow these steps:

  • Click NCERT Maths Class 10 Chapter 8 Exercise 8.4 Solutions.

  • You will be redirected to Vedantu’s Solutions page for Class 10 Maths Exercise 8.4.

  • At the top of the page, you will find the link for PDF for Exercise 8.4 Solutions.

  • Click on the link, and you can easily download the solutions and keep them with yourself in offline mode to practice whenever you want.

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