# Ex.8.2 Q2 Introduction to Trigonometry Solution - NCERT Maths Class 10

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## Question

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)  $${\rm{ sin}}\,{30^\circ}$$

Video Solution
Introduction To Trigonometry
Ex 8.2 | Question 2

## Text 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}\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.

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