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:

\(\begin{align}\rm{(i)} & \; \frac{{2\tan {{30}^\circ}}}{{1 + {{\tan }^2}\,{{30}^\circ}}}\\\\ & \left( {\rm{A}} \right){\rm{sin}}\,{60^\circ}\quad\left( {\rm{B}} \right){\rm{ cos}}\,{\rm{}}{60^\circ}{\rm{}}\quad\left( {\rm{C}} \right){\rm{ tan }}\,{60^\circ}{\rm{}}\quad\left( {\rm{D}} \right){\rm{ sin}}\,{60^\circ}\end{align}\)

\(\begin{align}\rm{(ii)} & \; \frac{{1 - {{\tan }^2}\,{{45}^\circ}}}{{1 + {{\tan }^2}\,{{45}^\circ}}}\\\\ & \left( {\rm{A}} \right){\rm{ tan}}\,{90^\circ}{\rm{}}\quad\left( {\rm{B}} \right){\rm{ 1}}\quad\left( {\rm{C}} \right){\rm{ sin}}\,{45^\circ}{\rm{}}\quad\left( {\rm{D}} \right){0^\circ}\end{align}\)

\(\begin{align}\rm{(iii)} & \; {\rm{sin}}\,2A = 2\,{\rm{sin}}\,A \quad {\text{ is true when A = }}\\\\ & \left( {\rm{A}} \right){0^\circ}\quad\left( {\rm{B}} \right){30^\circ}\quad\left( {\rm{C}} \right){45^\circ}\quad\left( {\rm{D}} \right){60^\circ}\end{align}\)

\(\begin{align}\rm{(iv)} & \; \frac{{2\tan {{30}^\circ}}}{{1 - {{\tan }^2}\,{{30}^\circ}}}\\\\ & \left( {\rm{A}} \right){\rm{ cos}}\,{\rm{}}{60^\circ}{\rm{}}\quad\left( {\rm{B}} \right){\rm{ sin}}\,{60^\circ}{\rm{}}\quad\left( {\rm{C}} \right){\rm{ tan}}\,{60^\circ}{\rm{}}\quad\left( {\rm{D}} \right){\rm{ sin}}\,{30^\circ}\end{align}\)

 

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

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

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

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

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

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

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

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

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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