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NCERT Solutions for Class 10 Maths Chapter 8 Introduction To Trigonometry Ex 8.2

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Exercise 8.2 Class 10 Maths NCERT Solutions for Chapter 8 - Free PDF Download

NCERT Solutions for Exercise 8.2 Class 10 Maths, Chapter 8, Introduction to Trigonometry, are available here with all the solved problems. It is available in free PDF format, which students can download and study with ease. The NCERT solutions are designed and reviewed by our subject matter experts with suitable diagrams, and the solutions are solved in a step-by-step manner. The solutions were created in accordance with the latest NCERT syllabus and guidelines of the CBSE board. All the solutions are given chapter-wise for easy access, helping students to solve the problems easily. The NCERT Class 10 Exercise 8.2 Maths Solutions for all the chapters can be downloaded at Vedantu for free.

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Glance on NCERT Solutions Maths Chapter 8 Ex 8.2 Class 10 | Vedantu

  • This chapter focuses on the formula of foundational trigonometric identities.

  • This chapter includes primarily theoretical questions that involve proving various identities and demonstrating the interrelationships between different trigonometric functions.

  • This chapter of Chapter 8 Maths Exercise 8.2 Class 10 helps students understand  Introduction to Trigonometry.

  • There are links to video tutorials explaining Chapter 8 Ex 8.2 Class 10 Introduction to Trigonometry for better understanding.

  • There are three exercises (19 fully solved questions) in Class 10 Maths, Chapter 8, Exercise 8.2  Introduction to Trigonometry.

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NCERT Solutions for Class 10 Maths Chapter 8 Introduction To Trigonometry Ex 8.2
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Access NCERT Solutions for Class -10 Maths Chapter 8 – Introduction to Trigonometry

Exercise 8.2

1. Evaluate the following:

(i)  $\sin 60{}^\circ \cos 30{}^\circ +\sin 30{}^\circ \cos 60{}^\circ $ 

Ans: With the help of trigonometric ratio tables we can find the values of standard trigonometric angles. The trigonometric ratio table is as follows:

Exact Values of Trigonometric Functions

Angle $\theta $ 

$\sin \theta $ 

$\cos \theta $ 

$\tan \theta $ 

Degrees

Radians

$0{}^\circ $ 

$0$ 

$0$

$1$ 

$0$

$30{}^\circ $ 

$\dfrac{\pi }{6}$ 

$\dfrac{1}{2}$ 

$\dfrac{\sqrt{3}}{2}$

$\dfrac{1}{\sqrt{3}}$ 

$45{}^\circ $ 

$\dfrac{\pi }{4}$ 

$\dfrac{1}{\sqrt{2}}$

$\dfrac{1}{\sqrt{2}}$ 

$1$ 

$60{}^\circ $ 

$\dfrac{\pi }{3}$ 

$\dfrac{\sqrt{3}}{2}$

$\dfrac{1}{2}$

$\sqrt{3}$

$90{}^\circ $ 

$\dfrac{\pi }{2}$ 

$1$ 

$0$

Not defined 


We have to evaluate $\sin 60{}^\circ \cos 30{}^\circ +\sin 30{}^\circ \cos 60{}^\circ $.

Substitute the values from the above table, we get

$\Rightarrow \dfrac{\sqrt{3}}{2}\times \dfrac{\sqrt{3}}{2}+\dfrac{1}{2}\times \dfrac{1}{2}$

$\Rightarrow \dfrac{3}{4}+\dfrac{1}{4}$ 

$\Rightarrow \dfrac{4}{4}$ 

$\therefore \sin 60{}^\circ \cos 30{}^\circ +\sin 30{}^\circ \cos 60{}^\circ =1$.


(ii)  $2{{\tan }^{2}}45{}^\circ +{{\cos }^{2}}30{}^\circ -{{\sin }^{2}}60{}^\circ $

Ans: With the help of trigonometric ratio tables we can find the values of standard trigonometric angles. The trigonometric ratio table is as follows:

Exact Values of Trigonometric Functions

Angle $\theta $ 

$\sin \theta $ 

$\cos \theta $ 

$\tan \theta $ 

Degrees

Radians

$0{}^\circ $ 

$0$ 

$0$

$1$ 

$0$

$30{}^\circ $ 

$\dfrac{\pi }{6}$ 

$\dfrac{1}{2}$ 

$\dfrac{\sqrt{3}}{2}$

$\dfrac{1}{\sqrt{3}}$ 

$45{}^\circ $ 

$\dfrac{\pi }{4}$ 

$\dfrac{1}{\sqrt{2}}$

$\dfrac{1}{\sqrt{2}}$ 

$1$ 

$60{}^\circ $ 

$\dfrac{\pi }{3}$ 

$\dfrac{\sqrt{3}}{2}$

$\dfrac{1}{2}$

$\sqrt{3}$

$90{}^\circ $ 

$\dfrac{\pi }{2}$ 

$1$ 

$0$

Not defined 


We have to evaluate $2{{\tan }^{2}}45{}^\circ +{{\cos }^{2}}30{}^\circ -{{\sin }^{2}}60{}^\circ $.

Substitute the values from the above table, we get

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

$\Rightarrow 2+\dfrac{3}{4}-\dfrac{3}{4}$ 

$\Rightarrow 2$ 

$\therefore 2{{\tan }^{2}}45{}^\circ +{{\cos }^{2}}30{}^\circ -{{\sin }^{2}}60{}^\circ =2$.


(iii)  $\dfrac{\cos 45{}^\circ }{\sec 30{}^\circ +\operatorname{cosec}30{}^\circ }$

Ans: With the help of trigonometric ratio tables we can find the values of standard trigonometric angles. The trigonometric ratio table is as follows:

Exact Values of Trigonometric Functions

Angle $\theta $ 

$\sin \theta $ 

$\cos \theta $ 

$\tan \theta $ 

Degrees

Radians

$0{}^\circ $ 

$0$ 

$0$

$1$ 

$0$

$30{}^\circ $ 

$\dfrac{\pi }{6}$ 

$\dfrac{1}{2}$ 

$\dfrac{\sqrt{3}}{2}$

$\dfrac{1}{\sqrt{3}}$ 

$45{}^\circ $ 

$\dfrac{\pi }{4}$ 

$\dfrac{1}{\sqrt{2}}$

$\dfrac{1}{\sqrt{2}}$ 

$1$ 

$60{}^\circ $ 

$\dfrac{\pi }{3}$ 

$\dfrac{\sqrt{3}}{2}$

$\dfrac{1}{2}$

$\sqrt{3}$

$90{}^\circ $ 

$\dfrac{\pi }{2}$ 

$1$ 

$0$

Not defined 


We have to evaluate $\dfrac{\cos 45{}^\circ }{\sec 30{}^\circ +\operatorname{cosec}30{}^\circ }$.

Substitute the values from the above table, we get

$\Rightarrow \dfrac{\dfrac{1}{\sqrt{2}}}{\dfrac{2}{\sqrt{3}}+2}$

$\Rightarrow \dfrac{\dfrac{1}{\sqrt{2}}}{\dfrac{2+2\sqrt{3}}{\sqrt{3}}}$ 

\[\Rightarrow \dfrac{1}{\sqrt{2}}\times \dfrac{\sqrt{3}}{2+2\sqrt{3}}\]

Multiplying and dividing by \[\sqrt{3}-1\], we get

\[\Rightarrow \dfrac{1}{\sqrt{2}}\times \dfrac{\sqrt{3}}{2+2\sqrt{3}}\times \dfrac{\sqrt{3}-1}{\sqrt{3}-1}\]

\[\Rightarrow \dfrac{\sqrt{3}\left( \sqrt{3}-1 \right)}{\sqrt{2}\left( 2+2\sqrt{3} \right)\left( \sqrt{3}-1 \right)}\]

\[\Rightarrow \dfrac{\sqrt{3}\left( \sqrt{3}-1 \right)}{2\sqrt{2}\left( \sqrt{3}+1 \right)\left( \sqrt{3}-1 \right)}\]

\[\Rightarrow \dfrac{3-\sqrt{3}}{2\sqrt{2}\left( {{\left( \sqrt{3} \right)}^{2}}-{{1}^{2}} \right)}\]

\[\Rightarrow \dfrac{3-\sqrt{3}}{2\sqrt{2}\left( 3-1 \right)}\]

$\Rightarrow \dfrac{3-\sqrt{3}}{4\sqrt{2}}$ 

$\therefore \dfrac{\cos 45{}^\circ }{\sec 30{}^\circ +\operatorname{cosec}30{}^\circ }=\dfrac{3-\sqrt{3}}{4\sqrt{2}}$

 

(iv)  $\dfrac{\sin 30{}^\circ +\tan 45{}^\circ -\operatorname{cosec}60{}^\circ }{\sec 30{}^\circ +\cos 60{}^\circ -\cot 45{}^\circ }$

Ans: With the help of trigonometric ratio tables we can find the values of standard trigonometric angles. The trigonometric ratio table is as follows:

Exact Values of Trigonometric Functions

Angle $\theta $ 

$\sin \theta $ 

$\cos \theta $ 

$\tan \theta $ 

Degrees

Radians

$0{}^\circ $ 

$0$ 

$0$

$1$ 

$0$

$30{}^\circ $ 

$\dfrac{\pi }{6}$ 

$\dfrac{1}{2}$ 

$\dfrac{\sqrt{3}}{2}$

$\dfrac{1}{\sqrt{3}}$ 

$45{}^\circ $ 

$\dfrac{\pi }{4}$ 

$\dfrac{1}{\sqrt{2}}$

$\dfrac{1}{\sqrt{2}}$ 

$1$ 

$60{}^\circ $ 

$\dfrac{\pi }{3}$ 

$\dfrac{\sqrt{3}}{2}$

$\dfrac{1}{2}$

$\sqrt{3}$

$90{}^\circ $ 

$\dfrac{\pi }{2}$ 

$1$ 

$0$

Not defined 

 

We have to evaluate $\dfrac{\sin 30{}^\circ +\tan 45{}^\circ -\operatorname{cosec}60{}^\circ }{\sec 30{}^\circ +\cos 60{}^\circ -\cot 45{}^\circ }$.

Substitute the values from the above table, we get

$\Rightarrow \dfrac{\dfrac{1}{2}+1-\dfrac{2}{\sqrt{3}}}{\dfrac{2}{\sqrt{3}}+\dfrac{1}{2}+1}$

$\Rightarrow \dfrac{\dfrac{3}{2}-\dfrac{2}{\sqrt{3}}}{\dfrac{2}{\sqrt{3}}+\dfrac{3}{2}}$

$\Rightarrow \dfrac{\dfrac{3\sqrt{3}-4}{2\sqrt{3}}}{\dfrac{3\sqrt{3}+4}{2\sqrt{3}}}$

$\Rightarrow \dfrac{3\sqrt{3}-4}{3\sqrt{3}+4}$

Multiplying and dividing by \[3\sqrt{3}-4\], we get

\[\Rightarrow \dfrac{3\sqrt{3}-4}{3\sqrt{3}+4}\times \dfrac{3\sqrt{3}-4}{3\sqrt{3}-4}\]

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

\[\Rightarrow \dfrac{{{\left( 3\sqrt{3}-4 \right)}^{2}}}{{{\left( 3\sqrt{3} \right)}^{2}}-{{4}^{2}}}\]

\[\Rightarrow \dfrac{{{\left( 3\sqrt{3}-4 \right)}^{2}}}{{{\left( 3\sqrt{3} \right)}^{2}}-{{4}^{2}}}\]

\[\Rightarrow \dfrac{27+16-24\sqrt{3}}{27-16}\]

$\Rightarrow \dfrac{43-24\sqrt{3}}{11}$ 

$\therefore \dfrac{\sin 30{}^\circ +\tan 45{}^\circ -\operatorname{cosec}60{}^\circ }{\sec 30{}^\circ +\cos 60{}^\circ -\cot 45{}^\circ }=\dfrac{43-24\sqrt{3}}{11}$


(v) $\dfrac{5{{\cos }^{2}}60{}^\circ +4{{\sec }^{2}}30{}^\circ -{{\tan }^{2}}45{}^\circ }{{{\sec }^{2}}30{}^\circ +{{\cos }^{2}}30{}^\circ }$

Ans: With the help of trigonometric ratio tables we can find the values of standard trigonometric angles. The trigonometric ratio table is as follows:

Exact Values of Trigonometric Functions

Angle $\theta $ 

$\sin \theta $ 

$\cos \theta $ 

$\tan \theta $ 

Degrees

Radians

$0{}^\circ $ 

$0$ 

$0$

$1$ 

$0$

$30{}^\circ $ 

$\dfrac{\pi }{6}$ 

$\dfrac{1}{2}$ 

$\dfrac{\sqrt{3}}{2}$

$\dfrac{1}{\sqrt{3}}$ 

$45{}^\circ $ 

$\dfrac{\pi }{4}$ 

$\dfrac{1}{\sqrt{2}}$

$\dfrac{1}{\sqrt{2}}$ 

$1$ 

$60{}^\circ $ 

$\dfrac{\pi }{3}$ 

$\dfrac{\sqrt{3}}{2}$

$\dfrac{1}{2}$

$\sqrt{3}$

$90{}^\circ $ 

$\dfrac{\pi }{2}$ 

$1$ 

$0$

Not defined 


We have to evaluate $\dfrac{5{{\cos }^{2}}60{}^\circ +4{{\sec }^{2}}30{}^\circ -{{\tan }^{2}}45{}^\circ }{{{\sec }^{2}}30{}^\circ +{{\cos }^{2}}30{}^\circ }$.

Substitute the values from the above table, we get

$\Rightarrow \dfrac{5{{\left( \dfrac{1}{2} \right)}^{2}}+4{{\left( \dfrac{2}{\sqrt{3}} \right)}^{2}}-{{1}^{2}}}{{{\left( \dfrac{1}{2} \right)}^{2}}+{{\left( \dfrac{\sqrt{3}}{2} \right)}^{2}}}$

$\Rightarrow \dfrac{5\left( \dfrac{1}{4} \right)+4\left( \dfrac{4}{3} \right)-1}{\left( \dfrac{1}{4} \right)+\left( \dfrac{3}{4} \right)}$

$\Rightarrow \dfrac{\dfrac{15+64-12}{12}}{\dfrac{1+3}{4}}$

$\Rightarrow \dfrac{\dfrac{15+64-12}{12}}{\dfrac{1+3}{4}}$

$\Rightarrow \dfrac{\dfrac{15+64-12}{12}}{\dfrac{4}{4}}$ 

$\Rightarrow \dfrac{\dfrac{67}{12}}{1}$ 

$\therefore \dfrac{5{{\cos }^{2}}60{}^\circ +4{{\sec }^{2}}30{}^\circ -{{\tan }^{2}}45{}^\circ }{{{\sec }^{2}}30{}^\circ +{{\cos }^{2}}30{}^\circ }=\dfrac{67}{12}$.


2. Choose the correct option and justify your choice.

(i)  $\dfrac{2\tan 30{}^\circ }{1+{{\tan }^{2}}30{}^\circ }=$ ………

  1. $\sin 60{}^\circ $ 

  2. $\cos 60{}^\circ $ 

  3. $\tan 60{}^\circ $ 

  4. $\sin 30{}^\circ $ 

Ans: The given expression is $\dfrac{2\tan 30{}^\circ }{1+{{\tan }^{2}}30{}^\circ }$.

We know that from the trigonometric ratio table we have $\tan 30{}^\circ =\dfrac{1}{\sqrt{3}}$.

Substitute the value in the given expression we get

$\dfrac{2\tan 30{}^\circ }{1+{{\tan }^{2}}30{}^\circ }=\dfrac{2\left( \dfrac{1}{\sqrt{3}} \right)}{1+{{\left( \dfrac{1}{\sqrt{3}} \right)}^{2}}}$

$\Rightarrow \dfrac{2\tan 30{}^\circ }{1+{{\tan }^{2}}30{}^\circ }=\dfrac{\dfrac{2}{\sqrt{3}}}{1+\dfrac{1}{3}}$

$\Rightarrow \dfrac{2\tan 30{}^\circ }{1+{{\tan }^{2}}30{}^\circ }=\dfrac{\dfrac{2}{\sqrt{3}}}{\dfrac{4}{3}}$

$\Rightarrow \dfrac{2\tan 30{}^\circ }{1+{{\tan }^{2}}30{}^\circ }=\dfrac{\sqrt{3}}{2}$

From the trigonometric table we know that 

$\sin 60{}^\circ =\dfrac{\sqrt{3}}{2}$ 

$\cos 60{}^\circ =\dfrac{1}{2}$ 

$\tan 60{}^\circ =\sqrt{3}$ 

$\sin 30{}^\circ =\dfrac{1}{2}$

Hence, $\dfrac{2\tan 30{}^\circ }{1+{{\tan }^{2}}30{}^\circ }=\sin 60{}^\circ $.

Therefore, option (A) is the correct answer.


(ii)  $\dfrac{1-{{\tan }^{2}}45{}^\circ }{1+{{\tan }^{2}}45{}^\circ }=$ ………

  1. $\tan 90{}^\circ $ 

  2. $1$ 

  3. $\sin 45{}^\circ $ 

  4. $0$

Ans: The given expression is $\dfrac{1-{{\tan }^{2}}45{}^\circ }{1+{{\tan }^{2}}45{}^\circ }$.

We know that from the trigonometric ratio table we have $\tan 45{}^\circ =1$.

Substitute the value in the given expression we get

$\dfrac{1-{{\tan }^{2}}45{}^\circ }{1+{{\tan }^{2}}45{}^\circ }=\dfrac{1-{{1}^{2}}}{1+{{1}^{2}}}$

$\Rightarrow \dfrac{1-{{\tan }^{2}}45{}^\circ }{1+{{\tan }^{2}}45{}^\circ }=\dfrac{1-1}{1+1}$

$\Rightarrow \dfrac{1-{{\tan }^{2}}45{}^\circ }{1+{{\tan }^{2}}45{}^\circ }=\dfrac{0}{2}$

$\therefore \dfrac{1-{{\tan }^{2}}45{}^\circ }{1+{{\tan }^{2}}45{}^\circ }=0$

Therefore, option (D) is the correct answer.


(iii)  $\sin 2A=2\sin A$ is true when $A=$ ……..

  1. $0{}^\circ $ 

  2. $30{}^\circ $ 

  3. $45{}^\circ $ 

  4. $60{}^\circ $ 

Ans: The given expression is $\sin 2A=2\sin A$.

We know that from the trigonometric ratio table we have 

$\sin 0{}^\circ =0$

$\sin 30{}^\circ =\dfrac{1}{2}$

$\sin 45{}^\circ =\dfrac{1}{\sqrt{2}}$

$\sin 60{}^\circ =\dfrac{\sqrt{3}}{2}$  

$\sin 90{}^\circ =1$ 

The given statement is true when $A=0{}^\circ $.

Substitute the value in the given expression we get

$\Rightarrow \sin 2A=2\sin A$

$\Rightarrow \sin 2\times 0{}^\circ =2\sin 0{}^\circ $

$0=0$ 

Therefore, option (A) is the correct answer.


(iv)  $\dfrac{2\tan 30{}^\circ }{1-{{\tan }^{2}}30{}^\circ }=$………

  1. $\sin 60{}^\circ $ 

  2. $\cos 60{}^\circ $ 

  3. $\tan 60{}^\circ $ 

  4. $\sin 30{}^\circ $ 

Ans: The given expression is $\dfrac{2\tan 30{}^\circ }{1-{{\tan }^{2}}30{}^\circ }$.

We know that from the trigonometric ratio table we have $\tan 30{}^\circ =\dfrac{1}{\sqrt{3}}$.

Substitute the value in the given expression we get

$\dfrac{2\tan 30{}^\circ }{1-{{\tan }^{2}}30{}^\circ }=\dfrac{2\left( \dfrac{1}{\sqrt{3}} \right)}{1-{{\left( \dfrac{1}{\sqrt{3}} \right)}^{2}}}$

$\Rightarrow \dfrac{2\tan 30{}^\circ }{1-{{\tan }^{2}}30{}^\circ }=\dfrac{\dfrac{2}{\sqrt{3}}}{1-\dfrac{1}{3}}$

$\Rightarrow \dfrac{2\tan 30{}^\circ }{1-{{\tan }^{2}}30{}^\circ }=\dfrac{\dfrac{2}{\sqrt{3}}}{\dfrac{2}{3}}$

$\Rightarrow \dfrac{2\tan 30{}^\circ }{1-{{\tan }^{2}}30{}^\circ }=\sqrt{3}$

From the trigonometric table we know that 

$\sin 60{}^\circ =\dfrac{\sqrt{3}}{2}$ 

$\cos 60{}^\circ =\dfrac{1}{2}$ 

$\tan 60{}^\circ =\sqrt{3}$ 

$\sin 30{}^\circ =\dfrac{1}{2}$

Hence, $\dfrac{2\tan 30{}^\circ }{1-{{\tan }^{2}}30{}^\circ }=\tan 60{}^\circ $.

Therefore, option (C) is the correct answer.


3. If $\tan \left( A+B \right)=\sqrt{3}$ and $\tan \left( A-B \right)=\dfrac{1}{\sqrt{3}}$, $0{}^\circ <A+B\le 90{}^\circ $. Find $A$ and $B$.

Ans: Given that $\tan \left( A+B \right)=\sqrt{3}$ and $\tan \left( A-B \right)=\dfrac{1}{\sqrt{3}}$.

From the trigonometric ratio table we know that $\tan 60{}^\circ =\sqrt{3}$ and $\tan 30{}^\circ =\dfrac{1}{\sqrt{3}}$.

Then we get

$\tan \left( A+B \right)=\sqrt{3}$

$\Rightarrow \tan \left( A+B \right)=\tan 60{}^\circ $

$\Rightarrow A+B=60{}^\circ $ ……….(1)

Also, $\tan \left( A-B \right)=\dfrac{1}{\sqrt{3}}$

$\Rightarrow \tan \left( A-B \right)=\tan 30{}^\circ $

$\Rightarrow A-B=30{}^\circ $ ……….(2)

Adding eq. (1) and (2), we get

$2A=90{}^\circ $

$\therefore A=45{}^\circ $ 

Substitute the obtained value in eq. (1), we get

$45{}^\circ +B=60{}^\circ $ 

$\Rightarrow B=60{}^\circ -45{}^\circ $ 

$\therefore B=15{}^\circ $ 

Therefore, the values of $A$ and $B$ is $45{}^\circ $ and $15{}^\circ $ respectively.


4. State whether the following are true or false. Justify your answer.

(i)  $\sin \left( A+B \right)=\sin A+\sin B$.

Ans: Let us assume $A=30{}^\circ $ and $B=60{}^\circ $.

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

$\sin \left( A+B \right)$

Substitute the assumed values in the LHS, we get

$\sin \left( A+B \right)=\sin \left( 30{}^\circ +60{}^\circ  \right)$

$\Rightarrow \sin \left( A+B \right)=\sin \left( 90{}^\circ  \right)$ 

From the trigonometric ratio table we know that $\sin 90{}^\circ =1$, we get

$\Rightarrow \sin \left( A+B \right)=1$

Now, let us consider the RHS of the given expression and substitute the values, we get

$\sin A+\sin B=\sin 30{}^\circ +\sin 60{}^\circ $

From the trigonometric ratio table we know that $\sin 30{}^\circ =\dfrac{1}{2}$ and $\sin 60{}^\circ =\dfrac{\sqrt{3}}{2}$, we get

$\Rightarrow \sin A+\sin B=\dfrac{1}{2}+\dfrac{\sqrt{3}}{2}$

$\Rightarrow \sin A+\sin B=\dfrac{1+\sqrt{3}}{2}$

Thus, $LHS\ne RHS$.

Therefore, the given statement is false.


(ii)  The value of $\sin \theta $ increases as $\theta $ increases. 

Ans: The value of sine from the trigonometric ratio table is as follows:

$\sin 0{}^\circ =0$

$\sin 30{}^\circ =\dfrac{1}{2}=0.5$

$\sin 45{}^\circ =\dfrac{1}{\sqrt{2}}=0.707$

$\sin 60{}^\circ =\dfrac{\sqrt{3}}{2}=0.866$  

$\sin 90{}^\circ =1$ 

Therefore, we can conclude that the value of $\sin \theta $ increases as $\theta $ increases. 

Therefore, the given statement is true.


(iii)  The value of $\cos \theta $ increases as $\theta $ increases. 

Ans: The value of cosine from the trigonometric ratio table is as follows:

$\cos 0{}^\circ =1$

$\cos 30{}^\circ =\dfrac{\sqrt{3}}{2}=0.866$

$\cos 45{}^\circ =\dfrac{1}{\sqrt{2}}=0.707$

$\cos 60{}^\circ =\dfrac{1}{2}=0.5$  

$\cos 90{}^\circ =0$ 

Therefore, we can conclude that the value of $\cos \theta $ decreases as $\theta $ increases. 

Therefore, the given statement is false.


(iv)  \[\sin \theta =\cos \theta \] for all values of \[\theta \].

Ans: The trigonometric ratio table is given as follows:

Exact Values of Trigonometric Functions

Angle $\theta $ 

$\sin \theta $ 

$\cos \theta $ 

$\tan \theta $ 

Degrees

Radians

$0{}^\circ $ 

$0$ 

$0$

$1$ 

$0$

$30{}^\circ $ 

$\dfrac{\pi }{6}$ 

$\dfrac{1}{2}$ 

$\dfrac{\sqrt{3}}{2}$

$\dfrac{1}{\sqrt{3}}$ 

$45{}^\circ $ 

$\dfrac{\pi }{4}$ 

$\dfrac{1}{\sqrt{2}}$

$\dfrac{1}{\sqrt{2}}$ 

$1$ 

$60{}^\circ $ 

$\dfrac{\pi }{3}$ 

$\dfrac{\sqrt{3}}{2}$

$\dfrac{1}{2}$

$\sqrt{3}$

$90{}^\circ $ 

$\dfrac{\pi }{2}$ 

$1$ 

$0$

Not defined 

From the above table we can conclude that \[\sin \theta =\cos \theta \] is true only for $\theta =45{}^\circ $.

\[\sin \theta =\cos \theta \] is not true for all values of $\theta $.

Therefore, the given statement is false.

 

(v)  $\cot A$ is not defined for $A=0{}^\circ $.

Ans: We know that $\cot A=\dfrac{\cos A}{\sin A}$ .

If $A=0{}^\circ $, then $\cot 0{}^\circ =\dfrac{\cos 0{}^\circ }{\sin 0{}^\circ }$

From trigonometric ratio table we get

$\sin 0{}^\circ =0$ and $\cos 0{}^\circ =1$

We get

$\cot 0{}^\circ =\dfrac{1}{0}$, which is undefined.

Therefore, the given statement is true.


Conclusion

For a thorough understanding of trigonometry fundamentals, NCERT Solutions for Ex 8.2 Class 10 Maths Chapter 8, "Introduction to Trigonometry," is an excellent resource. This chapter is essential because it establishes a number of ideas that are fundamental to further research in engineering, physics, and mathematics. In particular, students should concentrate on becoming skilled in trigonometric identities and ratios, as well as how to use them to solve problems that include heights and distances. It is necessary to fully understand these ideas in order to answer a variety of exam questions.


NCERT Solutions for Class 10 Maths Chapter 8 Exercises

Exercise

Number of Questions

Exercise 8.1

11 Questions and Solutions

Exercise 8.3

4 Questions and Solutions


CBSE Class 10 Maths Chapter 8 Other Study Materials


Chapter-Specific NCERT Solutions for Class 10 Maths

Given below are the chapter-wise NCERT Solutions for Class 10 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.



Study Resources for Class 10 Maths

For complete preparation of Maths for CBSE Class 10 board exams, check out the following links for different study materials available at Vedantu.

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FAQs on NCERT Solutions for Class 10 Maths Chapter 8 Introduction To Trigonometry Ex 8.2

1. What are the main trigonometric ratios introduced in NCERT Solutions for Class 10 Maths Chapter 8 Exercise 8.2?

The main trigonometric ratios covered in NCERT Solutions for Class 10 Maths Chapter 8 Exercise 8.2 are sin (sine), cos (cosine), tan (tangent), cosec (cosecant), sec (secant), and cot (cotangent), evaluated for standard angles such as 0°, 30°, 45°, 60°, and 90°, as per the CBSE 2025–26 syllabus.

2. How do you find the value of a trigonometric ratio using NCERT Solutions for Class 10 Maths Chapter 8?

To find the value of a trigonometric ratio in NCERT Solutions for Class 10 Maths Chapter 8, follow these steps:

  • Identify the required angle and function (e.g., sin 30°)
  • Refer to the standard ratios table
  • Substitute the established value (e.g., sin 30° = 1/2)
  • Solve as per the question, following CBSE guidelines

3. Why is understanding trigonometric identities important for Class 10 students?

Trigonometric identities are crucial as they allow students to simplify complex expressions, prove mathematical statements, and solve equations efficiently. Mastery of these identities in Class 10 forms the foundation for future problems involving heights, distances, and advanced mathematics topics.

4. How many questions are there in Exercise 8.2 of NCERT Solutions for Class 10 Maths Chapter 8?

Exercise 8.2 of NCERT Solutions for Class 10 Maths Chapter 8 contains four questions as per the latest CBSE textbook, including short answer, long answer, and multiple-choice types.

5. What are common mistakes students make when solving trigonometric ratio problems in Class 10?

Common errors include:

  • Confusing trigonometric values of standard angles
  • Using the incorrect identity or formula
  • Calculation errors during simplification
  • Neglecting to check angle units (degrees vs radians)
NCERT Solutions guide students to avoid these by showing stepwise methods.

6. In what scenarios does the trigonometric function become undefined in NCERT Solutions for Class 10 Maths Chapter 8?

Some trigonometric functions are undefined for specific angles. For example, tan 90° and cot 0° are undefined, as their denominator becomes zero. Awareness of such restrictions is emphasized in the CBSE curriculum.

7. What is the benefit of practicing all types of questions in Exercise 8.2 for board exams?

Practicing every question type—short, long, and MCQ—in Exercise 8.2 builds confidence, strengthens conceptual clarity, and prepares students for the variability seen in actual CBSE exams, improving overall scores in mathematics.

8. How can you apply trigonometric ratios from Class 10 Chapter 8 to real-life situations?

Trigonometric ratios are used to calculate unknown heights, distances, and angles in fields like engineering, navigation, surveying, and architecture, which the NCERT Solutions introduce for foundational understanding.

9. What are the exact values of sin, cos, and tan for standard angles as per NCERT Solutions Class 10 Maths Chapter 8?

The exact values as per NCERT Solutions for Class 10 Maths Chapter 8 are:

  • sin 0° = 0,  sin 30° = 1/2,  sin 45° = 1/√2,  sin 60° = √3/2,  sin 90° = 1
  • cos 0° = 1,  cos 30° = √3/2,  cos 45° = 1/√2,  cos 60° = 1/2,  cos 90° = 0
  • tan 0° = 0,  tan 30° = 1/√3,  tan 45° = 1,  tan 60° = √3,  tan 90° = undefined

10. What if two trigonometric expressions appear to be equal for some angles but not all? How do NCERT Solutions help clarify?

NCERT Solutions teach that trigonometric identities hold universally, while some expressions (like sin A + sin B = sin(A+B)) may work only for specific values. Example-based justifications in the solutions show differences between identities and mere coincidences, deepening conceptual understanding as required by the CBSE exam pattern.

11. Can questions from solved examples in Exercise 8.2 appear in the CBSE board exams?

Yes, CBSE board exams often use patterns similar to solved examples in NCERT Solutions for Class 10 Maths Chapter 8 Exercise 8.2, sometimes changing the numbers but keeping the core concept. Practicing examples ensures students are familiar with such trends.

12. How do you verify the correctness of answers in trigonometric calculations according to NCERT Solutions?

To verify correctness,

  • Cross-check trigonometric values with standard tables
  • Substitute angle values back into the original expression
  • Review calculations for arithmetic errors
NCERT Solutions provide stepwise strategies for this, improving answer accuracy.

13. Why is it important to learn the complementary angle relationships in trigonometry in Class 10?

Complementary angle relationships (like sin(90°–A) = cos A) are essential because they:

  • Simplify complicated questions
  • Help in application-level problems
  • Form the basis for higher-level trigonometry and coordinate geometry
NCERT Solutions highlight their repetitive use and practical importance as per CBSE standards.