# NCERT Solutions For Class 10 Maths Chapter 8 Exercise 8.2

Go back to  'Introduction to Trigonometry'

## Chapter 8 Ex.8.2 Question 1

Evaluate the following:

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

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

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

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

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

### Solution

#### Reasoning:

We know that,

 Exact Values of Trigonometric Functions Angle ($$\theta$$) sin ($$\theta$$) cos ($$\theta$$) tan ($$\theta$$) Degrees Radians $$0^{\circ}$$ $$0$$ $$0$$ $$1$$ $$0$$ $$30^{\circ}$$ \begin{align}\frac{\pi }{6}\end{align} \begin{align}\frac{1}{2}\end{align} \begin{align}\frac{{\sqrt 3 }}{2}\end{align} \begin{align}\frac{1}{{\sqrt 3 }}\end{align} $$45^{\circ}$$ \begin{align}\frac{\pi }{4}\end{align} \begin{align}\frac{1}{{\sqrt 2 }}\end{align} \begin{align}\frac{1}{{\sqrt 2 }}\end{align} $$1$$ $$60^{\circ}$$ \begin{align}\frac{\pi }{3}\end{align} \begin{align}\frac{{\sqrt 3 }}{2}\end{align} \begin{align}\frac{1}{2}\end{align} \begin{align}\sqrt 3 \end{align} $$90^{\circ}$$ \begin{align}\frac{\pi }{2}\end{align} $$1$$ $$0$$ Not Defined

#### Steps:

(i)

\begin{align}&\sin {60^\circ}\cos {30^\circ} + \sin {30^\circ}\cos {60^\circ}\\ &= \left( {\frac{{\sqrt 3 }}{2}} \right)\left( {\frac{{\sqrt 3 }}{2}} \right) + \left( {\frac{1}{2}} \right)\left( {\frac{1}{2}} \right)\\ &= \frac{3}{4} + \frac{1}{4} = \frac{{3 + 1}}{4}\\ &= \frac{4}{4} = 1\end{align}

(ii)

\begin{align}&2{\tan ^2}\,{45^\circ } + {\cos ^2}\,{30^\circ } - {\sin ^2}\,{60^\circ }\\& \! = \!\! 2{\left( \! {\tan {{45}^\circ }}\! \right)^2}\!\!+\!\! {\left( \! {\cos {{30}^\circ }} \!\right)^2}\!\! - \!\!{\left(\! {\sin{{60}^\circ }}\! \right)^2} \\& = 2{(1)^2} + {\left( {\frac{{\sqrt 3 }}{2}} \right)^2} - {\left( {\frac{{\sqrt 3 }}{2}} \right)^2}\\ &= 2 + \frac{3}{4} - \frac{3}{4}\\ &\quad\\ &= 2\end{align}

(iii)

\begin{align}&\frac{{\cos \,{{45}^0}}}{{\sec \,{{30}^0} + {\rm{cosec}}\,{{30}^0}}} \\ &= \frac{{\left( {\frac{1}{{\sqrt 2 }}} \right)}}{{\left( {\frac{2}{{\sqrt 3 }}} \right) + \left( {\frac{2}{1}} \right)}}\\ &= \frac{{\frac{1}{{\sqrt 2 }}}}{{\frac{{2 + 2\sqrt 3 }}{{\sqrt 3 }}}}\\&= \frac{{1 \times \sqrt 3 }}{{\sqrt 2 \times \left( {2 + 2\sqrt 3 } \right)}}\\ &= \frac{{\sqrt 3 }}{{2\sqrt 2 \left( {\sqrt 3 + 1} \right)}}\end{align}

Multiplying numerator and denominator by $$\sqrt 2 \left( {\sqrt 3 - 1} \right)$$ , we get

\begin{align} &= \frac{{\sqrt 3 }}{{2\sqrt 2 \;\left( {\sqrt 3 + 1} \right)}} \times \frac{{\sqrt 2 (\sqrt 3 - 1)}}{{\sqrt 2 (\sqrt 3 - 1)}}\\ &= \frac{{3\sqrt 2 - \sqrt 6 }}{{4\;\left( {3 - 1} \right)}}\\ &= \frac{{3\sqrt 2 - \sqrt 6 }}{8} \end{align}

(iv)

\begin{align} & \frac{{\sin \,{{30}^0} + \tan \,{{45}^0} - {\rm{cosec}}\,{{60}^0}}}{{\sec \,{{30}^0} + \cos \,{{60}^0} + \cot \,{{45}^0}}} \\ \\ &= \frac{{\frac{1}{2} + 1 - \frac{2}{{\sqrt 3 }}}}{{\frac{2}{{\sqrt 3 }} + \frac{1}{2} + 1}}\\ &= \frac{{\frac{3}{2} - \frac{2}{{\sqrt 3 }}}}{{\frac{2}{{\sqrt 3 }} + \frac{3}{2}}}\\ &= \frac{{\frac{{3\sqrt 3 - 4}}{{2\sqrt 3 }}}}{{\frac{{4 + 3\sqrt 3 }}{{2\sqrt 3 }}}}\\ &= \frac{{3\sqrt 3 - 4}}{{3\sqrt 3 + 4}} \end{align}

Multiplying numerator and denominator by $$\left( {3\sqrt 3 - 4} \right)$$ ,we get

\begin{align} &= \frac{{(3\sqrt 3 - 4)\;(3\sqrt 3 - 4)}}{{(3\sqrt 3 + 4)\;(3\sqrt 3 - 4)}} & \\ &= \frac{{27 + 16 - 24\sqrt 3 }}{{27 - 16}}\\ &= \frac{{43 - 24\sqrt 3 }}{{11}} \end{align}

(v)

\begin{align} & \frac{{5{{\cos }^2}{{60}^0} + 4{{\sec }^2}{{30}^0} - {{\tan }^2}{{45}^0}}}{{{{\sin }^2}{{30}^0} + {{\cos }^2}{{30}^0}}} \\ &= \frac{{ \!\! 5\! \times \! {{\left( {\frac{1}{2}} \right)}^2}\! +\! 4 \! \times \!{{\left( \! {\frac{2}{{\sqrt 3 }}} \right)}^2}\! - \!{{( - 1)}^2} }}{ \left( \frac{1}{2} \right)^2 + \left( \frac{\sqrt 3 }{2} \right)^2 }\\ &= \frac{{\left( {\frac{5}{4} + \frac{{16}}{3} - 1} \right)}}{{\left( {\frac{1}{4} + \frac{3}{4}} \right)}}\\ &= \frac{{\left( {\frac{{15 + 64 - 12}}{{12}}} \right)}}{{\left( {\frac{{3 + 1}}{4}} \right)}}\\ & = \frac{{\left( {\frac{{67}}{{12}}} \right)}}{{\left( {\frac{4}{4}} \right)}}\\ &= \frac{{67}}{{12}} \end{align}

## Chapter 8 Ex.8.2 Question 2

Choose the correct option and justify your choice:

(i) $$\; \frac{{2\tan {{30}^\circ}}}{{1 + {{\tan }^2}\,{{30}^\circ}}}$$

(A) $${\rm{sin}}\,{60^\circ}$$

(B) $${\rm{ cos}}\,{\rm{}}{60^\circ}$$

(C) $${\rm{ tan }}\,{60^\circ}$$

(D) $${\rm{ sin}}\,{60^\circ}$$

(ii) $$\; \frac{{1 - {{\tan }^2}\,{{45}^\circ}}}{{1 + {{\tan }^2}\,{{45}^\circ}}}$$

(A) $${\rm{ tan}}\,{90^\circ}{\rm{}}$$

(B) $${\rm{ 1}}$$

(C) $${\rm{ sin}}\,{45^\circ}$$

(D) $${0^\circ}$$

(iii) $${\rm{sin}}\,2A = 2\,{\rm{sin}}\,A$$  is true when $$A =$$

(A) $${0^\circ}$$

(B) $${30^\circ}$$

(C) $${45^\circ}$$

(D) $${60^\circ}$$

(iv) $$\; \frac{{2\tan {{30}^\circ}}}{{1 - {{\tan }^2}\,{{30}^\circ}}}$$

(A) $${\rm{ cos}}\,{60^\circ}$$

(B) $${\rm{ sin}}\,{60^\circ}$$

(C) $${\rm{ tan}}\,{60^\circ}$$

(D)  $${\rm{ sin}}\,{30^\circ}$$

#### Reasoning:

We know that,

 Exact Values of Trigonometric Functions Angle ($$\theta$$) sin ($$\theta$$) cos ($$\theta$$) tan ($$\theta$$) Degrees Radians $$0^{\circ}$$ $$0$$ $$0$$ $$1$$ $$0$$ $$30^{\circ}$$ \begin{align}\frac{\pi }{6}\end{align} \begin{align}\frac{1}{2}\end{align} \begin{align}\frac{{\sqrt 3 }}{2}\end{align} \begin{align}\frac{1}{{\sqrt 3 }}\end{align} $$45^{\circ}$$ \begin{align}\frac{\pi }{4}\end{align} \begin{align}\frac{1}{{\sqrt 2 }}\end{align} \begin{align}\frac{1}{{\sqrt 2 }}\end{align} $$1$$ $$60^{\circ}$$ \begin{align}\frac{\pi }{3}\end{align} \begin{align}\frac{{\sqrt 3 }}{2}\end{align} \begin{align}\frac{1}{2}\end{align} \begin{align}\sqrt 3 \end{align} $$90^{\circ}$$ \begin{align}\frac{\pi }{2}\end{align} $$1$$ $$0$$ Not Defined

#### Steps:

(i)

\begin{align}\frac{{2\tan {{30}^\circ}}}{{1 + {{\tan }^2}\,{{30}^\circ}}} \end{align}

By substituting the values of given trigonometric ratios in the above equation, we get.

\begin{align} &= \frac{{2 \times \left( {\frac{1}{{\sqrt 3 }}} \right)}}{{1 + {{\left( {\frac{1}{{\sqrt 3 }}} \right)}^2}}}\\ &= \frac{{2 \times \frac{1}{{\sqrt 3 }}}}{{1 + \frac{1}{3}}}\\ &= \frac{{\frac{2}{{\sqrt 3 }}}}{{\frac{4}{{\sqrt 3 }}}}\\ &= \frac{6}{{4\sqrt 3 }}\\ &= \frac{{\sqrt 3 }}{2}\end{align}

Out of the given options only \begin{align}\sin \,{60^\circ} = \frac{{\sqrt 3 }}{2}.\end{align} Hence, option (A) is correct.

(ii)

\begin{align}\frac{{1 - {{\tan }^2}\,{{45}^\circ}}}{{1 + {{\tan }^2}\,{{45}^\circ}}}\end{align}

By substituting the values of given trigonometric ratios for$$\,\tan \,{45^\circ}.$$

\begin{align} &= \frac{{1 - {{(1)}^2}}}{{1 + {{(1)}^2}}}\\ &= \frac{{1 - 1}}{{1 + 1}}\\ &= \frac{0}{2}\\ &= 0\end{align}

Hence, option (D) is correct.

(iii)

\begin{align}{\rm{sin}}\,{\rm{2A}} = 2\,{\rm{sin}}\,{\rm{A}}\end{align}

By substituting \begin{align}{\rm{A}} = {0^\circ},\;{30^\circ},\;{45^\circ}\;{\rm{and}}\;{60^\circ}\end{align} we get

For \begin{align}{\rm{A}} = {0^\circ}\end{align}

\begin{align}\sin \,{\rm{2A}} &= \rm{sin}\,2 \times {0^\circ}\\ &= \rm{\sin} \,{0^\circ}\\&= 0\\2\rm{\sin} \,A &= 2 \times \rm{sin}\,{0^\circ}\\ &= 2 \times \,{0^\circ}\\ &= 0\\\rm{\sin} \,2A &= 2\,\rm{\sin} \,A \quad (When\;A = {0^\circ})\end{align}

For \begin{align}{\rm{A}} = {30^\circ}\end{align}

\begin{align}\rm{\sin \,2A }&= {\rm{\sin}}\,2 \times {30^\circ}\\ &= \sin \,{60^\circ}\\ &= \frac{{\sqrt 3 }}{2}\\2\sin \,A &= 2 \times \rm{sin}\,{30^\circ}\\ &= 2 \times \,\frac{1}{2}\\ &= 1\\\sin \,2A &\ne 2\sin \,A \\ & (\rm{When}\;A = {30^\circ})\end{align}

For \begin{align}{\rm{A}} = {45^\circ}\end{align}

\begin{align}\sin \,{\rm{2A}} &= \rm{sin\,2} \times {45^\circ}\\ &= \sin \,{90^\circ}\\ &= 1\\2\sin \,A & = 2 \times \rm{sin\,}{45^\circ}\\ &= 2 \times \,\frac{1}{{\sqrt 2 }}\\ &= \sqrt 2 \\\sin \,{\rm{2A}} &\ne 2\sin \,{\rm{A}} \\ & (\rm{When}\;A = {45^\circ})\end{align}

For \begin{align}{\rm{A}} = {60^\circ}\end{align}

\begin{align}\rm{\sin} \,{\rm{2A}} &= \rm{sin}\,2 \times {60^\circ}\\ &= \rm{\sin} \,{120^\circ}\\ &= \frac{{\sqrt 3 }}{2}\\2\;\rm{\sin} \,A &= 2 \times \rm{\sin}\,{60^\circ}\\ &= 2 \times \,\frac{{\sqrt 3 }}{2} = \sqrt 3 \\\sin \,{\rm{2A}} &\ne 2\sin \,{\rm{A}} \\& (\rm{When}\;A = {60^\circ})\end{align}

Hence Option (A) is correct

(iv)

\begin{align}\frac{{2\tan {{30}^\circ}}}{{1 - {{\tan }^2}\,{{30}^\circ}}}\end{align}

By substituting the values of given trigonometric ratios for \begin{align}\tan \,{30^\circ}\end{align} , we get

\begin{align} &= \frac{{2 \times \left( {\frac{1}{{\sqrt 3 }}} \right)}}{{1 - {{\left( {\frac{1}{{\sqrt 3 }}} \right)}^2}}}\\ &= \frac{{\left( {\frac{2}{{\sqrt 3 }}} \right)}}{{\left( {1 - \frac{1}{3}} \right)}}\\ &= \frac{{\left( {\frac{2}{{\sqrt 3 }}} \right)}}{{\left( {\frac{2}{3}} \right)}}\\ &= \sqrt 3 \end{align}

Out of the given option only \begin{align}\tan \,{60^\circ} = \sqrt 3 \,.\end{align}

Hence option (C) is correct.

## Chapter 8 Ex.8.2 Question 3

If \begin{align} \;\;\tan \,\left( \text{A + B} \right) = \sqrt 3 \end{align} and

\begin{align} & \tan \,\left( \rm{A\,-\,B} \right) = \frac{1}{{\sqrt 3 }};\\ & {0^\circ} < \left( \text{A + B} \right) \le 9{0^\circ},\; \rm{A} > \rm{B}, \end{align}

find $$\text{A}$$ and $$\text{B.}$$

#### Steps:

Given that

\begin{align}{\rm{tan}}\;(\rm{A + B}) &= \sqrt 3 \\{\rm{And,}}\;{\rm{tan}}\,(\rm{A} - \rm{B}) &= \frac{1}{{\sqrt 3 }}\\{\rm{Since}}\;\tan {60^\circ} &= \sqrt 3 \;{\rm{and}} \\ \tan \,{30^\circ} & = \frac{1}{{\sqrt 3 }}\\\text{Therefore,} \\ \therefore {\rm{tan}}\;(\rm{A + B}) &= {\rm{tan}}\;{60^\circ}\\rm{A + B}) &= {60^\circ} \qquad \dots{\rm{(i)}}\\ \therefore {\rm{tan}}\;(\rm{A} - \rm{B}) &= \tan \,{30^\circ}\\(\rm{A} - \rm{B}) &= {30^\circ}\qquad \dots{\rm{(ii)}}\end{align} On adding both equations (i) and (ii), we obtain: \begin{align} \rm{A + B + A} - \rm{B} &= {60^\circ} + {30^\circ}\\2\rm{A} &= {90^\circ}\\ \rm{A} &= {45^\circ}\end{align} By substituting the value of \(A in equation (i) we obtain

\begin{align} \rm{A + B} &= {60^\circ}\\{45^\circ} + \rm{B} &= {60^\circ}\\ \rm{B} &= {60^\circ} - {45^\circ}\\&= {15^\circ}\end{align}

Therefore,

\begin{align}\angle \rm{A} = {45^\circ}\;{\rm{and}}\;\angle \rm{A} = {15^\circ}\;\left( \rm{A > B} \right)\end{align}

## Chapter 8 Ex.8.2 Question 4

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

(i) \begin{align}\;\;{\rm{sin}}\,\left( {A + B} \right) = {\rm{sin}}\,A + {\rm{sin}}\,B\,.\end{align}

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

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

(iv) \begin{align}{\rm{sin}}\,\theta = {\rm{cos}}\,\theta \end{align} for all values of $$\theta$$.

(v) $$\cot A$$ is not defined for $$A = 0°$$.

### Solution

#### Steps:

(i)  \begin{align}{\rm{sin}}\,\left( {A + B} \right) = {\rm{sin}}\,A + {\rm{sin}}\,B\,.\end{align}

For the purpose of verification, Let \begin{align}A = {30^\circ}\;{\rm{and}}\;B = {60^\circ}\end{align}

\begin{align}L.H.S\, & = \sin \,\left( {A + B} \right)\\ &= \sin \,\left( {{{30}^\circ} + {{60}^\circ}} \right)\\ &= \sin \,{90^\circ}\\ &= 1\end{align}

\begin{align}R.H.S\, & = \sin A + \sin B\\ &= \sin {30^\circ} + \sin {60^\circ}\\ &= \frac{1}{2} + \frac{{\sqrt 3 }}{2}\\ &= \frac{{1 + \sqrt 3 }}{2}\end{align}

Since, \begin{align} \;\;{\rm{sin}}\,\left( {A + B} \right) \ne {\rm{sin}}\,A + {\rm{sin}}\,B\,.\end{align}

Hence, the given statement is not true

(ii) The value of \begin{align}\sin \,\theta \end{align} increases from 0 to 1 as $$\theta$$ increases from \begin{align}{0^\circ}\;{\rm{to}}\;{90^\circ}\end{align}

\begin{align}\sin {0^\circ} &= 0\\\sin {30^\circ} &= \frac{1}{2} = 0.5\\\sin {45^\circ} &= \frac{1}{{\sqrt 2 }} = 0.707\\\sin {60^\circ} &= \frac{{\sqrt 3 }}{2} = 0.866\\\sin {90^\circ} &= 1\end{align}

Hence, the given statement is true.

(iii) The value of \begin{align}\cos \theta \end{align} decreases from 1 to 0 as $$\theta$$ increases from \begin{align}{0^\circ}\;{\rm{to}}\;{90^\circ}\end{align}

\begin{align}\cos {0^\circ} &= 1\\\cos {30^\circ} &= \frac{{\sqrt 3 }}{2} = 0.866\\\cos {45^\circ} &= \frac{1}{{\sqrt 2 }} = 0.707\\\cos {60^\circ} &= \frac{1}{2} = 0.5\\\cos {90^\circ} &= 0\end{align}

Hence, the given statement is false.

(iv)

This is true when \begin{align}\theta = {45^\circ}\end{align}

As \begin{align}\sin {45^\circ} = \frac{1}{{\sqrt 2 }}\;\;{\rm{and}}\;\;\cos {45^\circ} = \frac{1}{{\sqrt 2 }}\end{align}

It is not true for other values of $$\theta$$

As,

\begin{align}\sin {30^\circ} &= \frac{1}{2}\quad{\rm{and}}\;\;\cos {30^\circ} = \frac{{\sqrt 3 }}{2}\\\sin {60^\circ} &= \frac{{\sqrt 3 }}{2}\quad{\rm{and}} \;\;\cos {60^\circ} \!\!= \!\! \frac{1}{{\sqrt 2 }}\\\sin {90^\circ} &= 1\quad{\rm{and}}\quad\cos {90^\circ} = 0\end{align}

Hence, the given statement is false.

(v)

\begin{align}\cot A &= \frac{{\cos A}}{{\sin A}}\\∴\; \cot {0^\circ} &= \frac{{\cos {0^\circ}}}{{\sin {0^\circ}}} \\ &= \frac{1}{0}\\ & = \text{undefined}\end{align}

Hence the given statement is true.

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