NCERT Solutions For Class 12 Maths Miscellaneous Exercise Chapter 2 Inverse Trigonometric Functions - 2025-26

Miscellaneous Solutions

1. Evaluate \[\text{co}{{\text{s}}^{\text{-1}}}\left( \text{cos}\dfrac{\text{13 }\!\!\pi\!\!\text{ }}{\text{6}} \right)\] 

Ans: Consider $\text{cos}\dfrac{\text{13 }\!\!\pi\!\!\text{ }}{\text{6}}$

$ \text{cos}\dfrac{\text{13 }\!\!\pi\!\!\text{ }}{\text{6}} $ 

$ \text{=cos}\left( \text{2 }\!\!\pi\!\!\text{ +}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}} \right) $ 

$ \text{=cos}\left( \dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}} \right) $ 

Further solving,

$ \text{co}{{\text{s}}^{\text{-1}}}\left( \text{cos}\dfrac{\text{13 }\!\!\pi\!\!\text{ }}{\text{6}} \right) $ 

$ \text{=co}{{\text{s}}^{\text{-1}}}\left( \text{cos}\left( \dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}} \right) \right) $ 

$\text{=}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}} $ 

Therefore, $\text{co}{{\text{s}}^{\text{-1}}}\left( \text{cos}\dfrac{\text{13 }\!\!\pi\!\!\text{ }}{\text{6}} \right)\text{=}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}}$

2. Evaluate \[\text{ta}{{\text{n}}^{\text{-1}}}\left( \text{tan}\dfrac{\text{7 }\!\!\pi\!\!\text{ }}{\text{6}} \right)\] 

Ans: Consider $\text{tan}\dfrac{\text{7 }\!\!\pi\!\!\text{ }}{\text{6}}$

$ \text{tan}\dfrac{\text{7 }\!\!\pi\!\!\text{ }}{\text{6}} $

$ \text{=tan}\left( \text{ }\!\!\pi\!\!\text{ +}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}} \right) $

$ \text{=tan}\left( \dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}} \right) $

Further solving,

$ \text{ta}{{\text{n}}^{\text{-1}}}\left( \text{tan}\dfrac{\text{7 }\!\!\pi\!\!\text{ }}{\text{6}} \right) $

$ \text{=ta}{{\text{n}}^{\text{-1}}}\left( \text{tan}\left( \dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}} \right) \right) $

$ \text{=}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}} $

Therefore, $\text{ta}{{\text{n}}^{\text{-1}}}\left( \text{tan}\dfrac{\text{7 }\!\!\pi\!\!\text{ }}{\text{6}} \right)\text{=}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}}$

3. Using inverse trigonometric identities, prove that $\text{2si}{{\text{n}}^{\text{-1}}}\dfrac{\text{3}}{\text{5}}\text{=ta}{{\text{n}}^{\text{-1}}}\dfrac{\text{24}}{\text{7}}$

Ans: Assuming \[\text{2si}{{\text{n}}^{\text{-1}}}\dfrac{\text{3}}{\text{5}}\text{= }\!\!\alpha\!\!\text{ }\] ----(1)

$ \text{sin}\dfrac{\text{ }\!\!\alpha\!\!\text{ }}{\text{2}}\text{=}\dfrac{\text{3}}{\text{5}} $ 

$ \text{cos}\dfrac{\text{ }\!\!\alpha\!\!\text{ }}{\text{2}}\text{=}\sqrt{\text{1-}{{\left( \dfrac{\text{3}}{\text{5}} \right)}^{\text{2}}}} $ 

 $ \text{cos}\dfrac{\text{ }\!\!\alpha\!\!\text{ }}{\text{2}}\text{=}\dfrac{\text{4}}{\text{5}} $ 

Hence,

$ \text{sin }\!\!\alpha\!\!\text{ =2}\left( \dfrac{\text{3}}{\text{5}} \right)\left( \dfrac{\text{4}}{\text{5}} \right) $ 

$ \text{sin }\!\!\alpha\!\!\text{ =}\dfrac{\text{24}}{\text{25}} $ 

$ \text{cos }\!\!\alpha\!\!\text{ =}\sqrt{\text{1-}{{\left( \dfrac{\text{24}}{\text{25}} \right)}^{\text{2}}}} $ 

$ \text{cos }\!\!\alpha\!\!\text{ =}\dfrac{\text{7}}{\text{25}} $ 

Therefore, 

\[\text{tan }\!\!\alpha\!\!\text{ =}\dfrac{\text{24}}{\text{7}}\]

\[\text{ }\!\!\alpha\!\!\text{ =ta}{{\text{n}}^{\text{-1}}}\left( \dfrac{\text{24}}{\text{7}} \right)\]

From Equation (1), it is proved that 

\[\text{2si}{{\text{n}}^{\text{-1}}}\dfrac{\text{3}}{\text{5}}\text{=ta}{{\text{n}}^{\text{-1}}}\left( \dfrac{\text{24}}{\text{7}} \right)\]

4. Using inverse trigonometric identities, prove that $\text{si}{{\text{n}}^{\text{-1}}}\dfrac{\text{8}}{\text{17}}\text{+si}{{\text{n}}^{\text{-1}}}\dfrac{\text{3}}{\text{5}}\text{=ta}{{\text{n}}^{\text{-1}}}\dfrac{\text{77}}{\text{36}}$

Ans: Assuming \[\text{si}{{\text{n}}^{\text{-1}}}\dfrac{\text{8}}{\text{17}}\text{= }\!\!\alpha\!\!\text{ }\] 

$\text{sin }\!\!\alpha\!\!\text{ =}\dfrac{\text{8}}{\text{17}} $ 

$ \text{cos }\!\!\alpha\!\!\text{ =}\sqrt{\text{1-}{{\left( \dfrac{\text{8}}{\text{17}} \right)}^{\text{2}}}} $ 

$ \text{cos }\!\!\alpha\!\!\text{ =}\dfrac{\text{15}}{\text{17}} $ 

Hence,

Therefore, 

$ \text{tan }\!\!\alpha\!\!\text{ =}\dfrac{\text{8}}{\text{15}} $ 

$ \text{ }\!\!\alpha\!\!\text{ =ta}{{\text{n}}^{\text{-1}}}\left( \dfrac{\text{8}}{\text{15}} \right) $ 

\[\text{si}{{\text{n}}^{\text{-1}}}\left( \dfrac{\text{8}}{\text{17}} \right)\text{=ta}{{\text{n}}^{\text{-1}}}\left( \dfrac{\text{8}}{\text{15}} \right)\] ----(1) 

Assuming that \[\text{si}{{\text{n}}^{\text{-1}}}\frac{\text{3}}{\text{5}}\text{= }\!\!\beta\!\!\text{ }\]

$ \text{sin }\!\!\beta\!\!\text{ =}\dfrac{\text{3}}{\text{5}} $ 

$ \text{cos }\!\!\beta\!\!\text{ =}\sqrt{\text{1-}{{\left( \dfrac{\text{3}}{\text{5}} \right)}^{\text{2}}}} $ 

$\text{cos }\!\!\beta\!\!\text{ =}\dfrac{\text{4}}{\text{5}} $ 

Therefore, 

$ \text{tan }\!\!\beta\!\!\text{ =}\dfrac{\text{3}}{\text{4}} $ 

$ \text{ }\!\!\beta\!\!\text{ =ta}{{\text{n}}^{\text{-1}}}\left( \dfrac{\text{3}}{\text{4}} \right) $ 

\[\text{si}{{\text{n}}^{\text{-1}}}\left( \dfrac{\text{3}}{\text{5}} \right)\text{=ta}{{\text{n}}^{\text{-1}}}\left( \dfrac{\text{3}}{\text{4}} \right)\] ----(2) 

Consider 

\[\text{si}{{\text{n}}^{\text{-1}}}\left( \dfrac{\text{8}}{\text{17}} \right)\text{+si}{{\text{n}}^{\text{-1}}}\left( \dfrac{\text{3}}{\text{5}} \right)\]

From Equations (1) and (2),

$ \text{si}{{\text{n}}^{\text{-1}}}\left( \dfrac{\text{8}}{\text{17}} \right)\text{+si}{{\text{n}}^{\text{-1}}}\left( \dfrac{\text{3}}{\text{5}} \right) $ 

$ \text{=ta}{{\text{n}}^{\text{-1}}}\left( \dfrac{\text{8}}{\text{15}} \right)\text{+ta}{{\text{n}}^{\text{-1}}}\left( \dfrac{\text{3}}{\text{4}} \right) $ 

$ \text{=ta}{{\text{n}}^{\text{-1}}}\left( \dfrac{\dfrac{\text{8}}{\text{15}}\text{+}\dfrac{\text{3}}{\text{4}}}{\text{1-}\left( \dfrac{\text{8}}{\text{15}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{3}}{\text{4}} \right)} \right) $ 

$ \text{=ta}{{\text{n}}^{\text{-1}}}\left( \dfrac{\text{77}}{\text{36}} \right) $ 

Hence, it is proved that \[\text{si}{{\text{n}}^{\text{-1}}}\dfrac{\text{8}}{\text{17}}\text{+si}{{\text{n}}^{\text{-1}}}\dfrac{\text{3}}{\text{5}}\text{=ta}{{\text{n}}^{\text{-1}}}\dfrac{\text{77}}{\text{36}}\]

5. Using inverse trigonometric identities, prove that $\text{co}{{\text{s}}^{\text{-1}}}\dfrac{\text{4}}{\text{5}}\text{+co}{{\text{s}}^{\text{-1}}}\dfrac{\text{12}}{\text{13}}\text{=co}{{\text{s}}^{\text{-1}}}\dfrac{\text{33}}{\text{65}}$

Ans: Assuming \[\text{co}{{\text{s}}^{\text{-1}}}\dfrac{\text{4}}{\text{5}}\text{= }\!\!\alpha\!\!\text{ }\] 

$\text{cos }\!\!\alpha\!\!\text{ =}\dfrac{\text{4}}{\text{5}} $ 

$ \text{sin }\!\!\alpha\!\!\text{ =}\sqrt{\text{1-}{{\left( \dfrac{\text{4}}{\text{5}} \right)}^{\text{2}}}} $ 

$ \text{sin }\!\!\alpha\!\!\text{ =}\dfrac{\text{3}}{\text{5}} $ 

Therefore, 

$ \text{tan }\!\!\alpha\!\!\text{ =}\dfrac{\text{3}}{\text{4}} $ 

 $ \text{ }\!\!\alpha\!\!\text{ =ta}{{\text{n}}^{\text{-1}}}\left( \dfrac{\text{3}}{\text{4}} \right) $ 

\[\text{co}{{\text{s}}^{\text{-1}}}\left( \dfrac{\text{4}}{\text{5}} \right)\text{=ta}{{\text{n}}^{\text{-1}}}\left( \dfrac{\text{3}}{\text{4}} \right)\] ----(1) 

Assuming that \[\text{co}{{\text{s}}^{\text{-1}}}\dfrac{\text{12}}{\text{13}}\text{= }\!\!\beta\!\!\text{ }\]

$ \text{cos }\!\!\beta\!\!\text{ =}\dfrac{\text{12}}{\text{13}} $ 

$ \text{sin }\!\!\beta\!\!\text{ =}\sqrt{\text{1-}{{\left( \dfrac{\text{12}}{\text{13}} \right)}^{\text{2}}}} $ 

 $ \text{sin }\!\!\beta\!\!\text{ =}\dfrac{\text{5}}{\text{13}} $ 

Therefore, 

$ \text{tan }\!\!\beta\!\!\text{ =}\dfrac{\text{5}}{\text{12}} $ 

$ \text{ }\!\!\beta\!\!\text{ =ta}{{\text{n}}^{\text{-1}}}\left( \dfrac{\text{5}}{\text{12}} \right) $ 

\[\text{co}{{\text{s}}^{\text{-1}}}\left( \dfrac{\text{12}}{\text{13}} \right)\text{=ta}{{\text{n}}^{\text{-1}}}\left( \dfrac{\text{5}}{\text{12}} \right)\] ----(2) 

Consider 

\[\text{co}{{\text{s}}^{\text{-1}}}\left( \dfrac{\text{4}}{\text{5}} \right)\text{+co}{{\text{s}}^{\text{-1}}}\left( \dfrac{\text{12}}{\text{13}} \right)\]

From Equations (1) and (2),

$\text{co}{{\text{s}}^{\text{-1}}}\left( \dfrac{\text{4}}{\text{5}} \right)\text{+co}{{\text{s}}^{\text{-1}}}\left( \dfrac{\text{12}}{\text{13}} \right) $ 

$ \text{=ta}{{\text{n}}^{\text{-1}}}\left( \dfrac{\text{3}}{\text{4}} \right)\text{+ta}{{\text{n}}^{\text{-1}}}\left( \dfrac{\text{5}}{\text{12}} \right) $ 

$ \text{=ta}{{\text{n}}^{\text{-1}}}\left( \dfrac{\dfrac{\text{3}}{\text{4}}\text{+}\dfrac{\text{5}}{\text{12}}}{\text{1-}\left( \dfrac{\text{3}}{\text{4}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{5}}{\text{12}} \right)} \right) $ 

$ \text{=ta}{{\text{n}}^{\text{-1}}}\left( \dfrac{\text{56}}{\text{33}} \right) $ 

$ \text{=co}{{\text{s}}^{\text{-1}}}\left( \dfrac{\text{33}}{\text{65}} \right) $ 

Hence, it is proved that \[\text{co}{{\text{s}}^{\text{-1}}}\dfrac{\text{4}}{\text{5}}\text{+co}{{\text{s}}^{\text{-1}}}\dfrac{\text{12}}{\text{13}}\text{=co}{{\text{s}}^{\text{-1}}}\dfrac{\text{33}}{\text{65}}\]

6. Using inverse trigonometric identities, prove that $\text{co}{{\text{s}}^{\text{-1}}}\dfrac{\text{12}}{\text{13}}\text{+si}{{\text{n}}^{\text{-1}}}\dfrac{\text{3}}{\text{5}}\text{=si}{{\text{n}}^{\text{-1}}}\dfrac{\text{56}}{\text{65}}$

Ans: Assuming \[\text{co}{{\text{s}}^{\text{-1}}}\dfrac{\text{12}}{\text{13}}\text{= }\!\!\alpha\!\!\text{ }\] 

$ \text{cos }\!\!\alpha\!\!\text{ =}\dfrac{\text{12}}{\text{13}} $ 

$ \text{sin }\!\!\alpha\!\!\text{ =}\sqrt{\text{1-}{{\left( \dfrac{\text{12}}{\text{13}} \right)}^{\text{2}}}} $ 

$\text{sin }\!\!\alpha\!\!\text{ =}\frac{\text{5}}{\text{13}} $ 

Therefore, 

$ \text{tan }\!\!\alpha\!\!\text{ =}\frac{\text{5}}{\text{12}} $ 

$ \text{ }\!\!\alpha\!\!\text{ =ta}{{\text{n}}^{\text{-1}}}\left( \frac{\text{5}}{\text{12}} \right) $ 

\[\text{co}{{\text{s}}^{\text{-1}}}\left( \frac{\text{12}}{\text{13}} \right)\text{=ta}{{\text{n}}^{\text{-1}}}\left( \frac{\text{5}}{\text{12}} \right)\] ----(1) 

Assuming that \[\text{si}{{\text{n}}^{\text{-1}}}\frac{\text{3}}{\text{5}}\text{= }\!\!\beta\!\!\text{ }\]

$\text{sin }\!\!\beta\!\!\text{ =}\frac{\text{3}}{\text{5}} $ 

$ \text{cos }\!\!\beta\!\!\text{ =}\sqrt{\text{1-}{{\left( \frac{\text{3}}{\text{5}} \right)}^{\text{2}}}} $ 

$ \text{cos }\!\!\beta\!\!\text{ =}\frac{\text{4}}{\text{5}} $ 

Therefore, 

$\text{tan }\!\!\beta\!\!\text{ =}\frac{\text{3}}{\text{4}} $ 

$ \text{ }\!\!\beta\!\!\text{ =ta}{{\text{n}}^{\text{-1}}}\left( \frac{\text{3}}{\text{4}} \right) $ 

\[\text{si}{{\text{n}}^{\text{-1}}}\left( \frac{\text{3}}{\text{5}} \right)\text{=ta}{{\text{n}}^{\text{-1}}}\left( \frac{\text{3}}{\text{4}} \right)\] ----(2) 

Consider 

\[\text{co}{{\text{s}}^{\text{-1}}}\left( \frac{\text{12}}{\text{13}} \right)\text{+si}{{\text{n}}^{\text{-1}}}\left( \frac{\text{3}}{\text{5}} \right)\]

From Equations (1) and (2),

$ \text{co}{{\text{s}}^{\text{-1}}}\left( \frac{\text{12}}{\text{13}} \right)\text{+si}{{\text{n}}^{\text{-1}}}\left( \frac{\text{3}}{\text{5}} \right) $ 

$ \text{=ta}{{\text{n}}^{\text{-1}}}\left( \frac{\text{5}}{\text{12}} \right)\text{+ta}{{\text{n}}^{\text{-1}}}\left( \frac{\text{3}}{\text{4}} \right) $ 

$ \text{=ta}{{\text{n}}^{\text{-1}}}\left( \frac{\frac{\text{3}}{\text{4}}\text{+}\frac{\text{5}}{\text{12}}}{\text{1-}\left( \frac{\text{3}}{\text{4}}\text{ }\!\!\times\!\!\text{ }\frac{\text{5}}{\text{12}} \right)} \right) $ 

$ \text{=ta}{{\text{n}}^{\text{-1}}}\left( \frac{\text{56}}{\text{33}} \right) $ 

$ \text{=si}{{\text{n}}^{\text{-1}}}\left( \frac{\text{56}}{\text{65}} \right) $ 

Hence, it is proved that \[\text{co}{{\text{s}}^{\text{-1}}}\frac{\text{12}}{\text{13}}\text{+si}{{\text{n}}^{\text{-1}}}\frac{\text{3}}{\text{5}}\text{=si}{{\text{n}}^{\text{-1}}}\frac{\text{56}}{\text{65}}\]

7. Using inverse trigonometric identities, prove that $\text{ta}{{\text{n}}^{\text{-1}}}\frac{\text{63}}{\text{16}}\text{=si}{{\text{n}}^{\text{-1}}}\frac{\text{5}}{\text{13}}\text{+co}{{\text{s}}^{\text{-1}}}\frac{\text{3}}{\text{5}}$

Ans: Assuming \[\text{si}{{\text{n}}^{\text{-1}}}\frac{\text{5}}{\text{13}}\text{= }\!\!\alpha\!\!\text{ }\] 

$ \text{sin }\!\!\alpha\!\!\text{ =}\frac{\text{5}}{\text{13}} $ 

$ \text{cos }\!\!\alpha\!\!\text{ =}\sqrt{\text{1-}{{\left( \frac{\text{5}}{\text{13}} \right)}^{\text{2}}}} $ 

$ \text{cos }\!\!\alpha\!\!\text{ =}\frac{\text{12}}{\text{13}} $ 

Therefore, 

$ \text{tan }\!\!\alpha\!\!\text{ =}\frac{\text{5}}{\text{12}} $ 

$ \text{ }\!\!\alpha\!\!\text{ =ta}{{\text{n}}^{\text{-1}}}\left( \frac{\text{5}}{\text{12}} \right) $ 

\[\text{si}{{\text{n}}^{\text{-1}}}\left( \frac{\text{5}}{\text{13}} \right)\text{=ta}{{\text{n}}^{\text{-1}}}\left( \frac{\text{5}}{\text{12}} \right)\] ----(1) 

Assuming that \[\text{co}{{\text{s}}^{\text{-1}}}\frac{\text{3}}{\text{5}}\text{= }\!\!\beta\!\!\text{ }\]

$ \text{cos }\!\!\beta\!\!\text{ =}\frac{\text{3}}{\text{5}} $ 

$ \text{sin }\!\!\beta\!\!\text{ =}\sqrt{\text{1-}{{\left( \frac{\text{3}}{\text{5}} \right)}^{\text{2}}}} $ 

$ \text{sin }\!\!\beta\!\!\text{ =}\frac{\text{4}}{\text{5}} $ 

Therefore, 

$\text{tan }\!\!\beta\!\!\text{ =}\frac{\text{4}}{\text{3}} $ 

$\text{ }\!\!\beta\!\!\text{ =ta}{{\text{n}}^{\text{-1}}}\left( \frac{\text{4}}{\text{3}} \right) $ 

\[\text{co}{{\text{s}}^{\text{-1}}}\left( \frac{\text{3}}{\text{5}} \right)\text{=ta}{{\text{n}}^{\text{-1}}}\left( \frac{\text{4}}{\text{3}} \right)\] ----(2) 

Consider 

\[\text{si}{{\text{n}}^{\text{-1}}}\left( \frac{\text{5}}{\text{13}} \right)\text{+co}{{\text{s}}^{\text{-1}}}\left( \frac{\text{3}}{\text{5}} \right)\]

From Equations (1) and (2),

$ \text{si}{{\text{n}}^{\text{-1}}}\left( \frac{\text{5}}{\text{13}} \right)\text{+co}{{\text{s}}^{\text{-1}}}\left( \frac{\text{3}}{\text{5}} \right) $ 

$ \text{=ta}{{\text{n}}^{\text{-1}}}\left( \frac{\text{5}}{\text{12}} \right)\text{+ta}{{\text{n}}^{\text{-1}}}\left( \frac{\text{4}}{\text{3}} \right) $ 

 $ \text{=ta}{{\text{n}}^{\text{-1}}}\left( \frac{\frac{\text{4}}{\text{3}}\text{+}\frac{\text{5}}{\text{12}}}{\text{1-}\left( \frac{\text{4}}{\text{3}}\text{ }\!\!\times\!\!\text{ }\frac{\text{5}}{\text{12}} \right)} \right) $ 

$ \text{=ta}{{\text{n}}^{\text{-1}}}\left( \frac{\text{63}}{\text{16}} \right) $ 

Hence, it is proved that \[\text{ta}{{\text{n}}^{\text{-1}}}\frac{\text{63}}{\text{16}}\text{=si}{{\text{n}}^{\text{-1}}}\frac{\text{5}}{\text{13}}\text{+co}{{\text{s}}^{\text{-1}}}\frac{\text{3}}{\text{5}}\]

8. Using inverse trigonometric identities, prove that ${{\tan }^{-1}}\sqrt{x}=\frac{1}{2}{{\cos }^{-1}}\left( \frac{1-x}{1+x} \right),x\in \left[ 0,1 \right]$

Ans: Assuming \[\text{x=ta}{{\text{n}}^{\text{2}}}\text{ }\!\!\alpha\!\!\text{ }\] 

Consider

$\text{ta}{{\text{n}}^{\text{-1}}}\sqrt{\text{x}} $ 

 $ \text{=ta}{{\text{n}}^{\text{-1}}}\sqrt{\text{ta}{{\text{n}}^{\text{2}}}\text{ }\!\!\alpha\!\!\text{ }} $ 

$ \text{=ta}{{\text{n}}^{\text{-1}}}\text{tan }\!\!\alpha\!\!\text{ } $ 

$\text{= }\!\!\alpha\!\!\text{ } $ 

\[\text{ta}{{\text{n}}^{\text{-1}}}\sqrt{\text{x}}\text{= }\!\!\alpha\!\!\text{ }\]----(1)

Consider 

$\frac{\text{1}}{\text{2}}\text{co}{{\text{s}}^{\text{-1}}}\left( \frac{\text{1-x}}{\text{1+x}} \right) $ 

$ \text{=}\frac{\text{1}}{\text{2}}\text{co}{{\text{s}}^{\text{-1}}}\left( \frac{\text{1-ta}{{\text{n}}^{\text{2}}}\text{ }\!\!\alpha\!\!\text{ }}{\text{1+ta}{{\text{n}}^{\text{2}}}\text{ }\!\!\alpha\!\!\text{ }} \right) $ 

$ \text{=}\frac{\text{1}}{\text{2}}\text{co}{{\text{s}}^{\text{-1}}}\text{cos2 }\!\!\alpha\!\!\text{ } $ 

$ \text{= }\!\!\alpha\!\!\text{ } $ 

\[\frac{\text{1}}{\text{2}}\text{co}{{\text{s}}^{\text{-1}}}\left( \frac{\text{1-x}}{\text{1+x}} \right)\text{= }\!\!\alpha\!\!\text{ }\] -----(2)

From Equations (1) and (2),

It is proved that \[\text{ta}{{\text{n}}^{\text{-1}}}\sqrt{\text{x}}\text{=}\frac{\text{1}}{\text{2}}\text{co}{{\text{s}}^{\text{-1}}}\left( \frac{\text{1-x}}{\text{1+x}} \right)\]

9. Using inverse trigonometric identities, prove that ${{\cot }^{-1}}\left( \frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}} \right)=\frac{x}{2},x\in \left( 0,\frac{\pi }{4} \right)$

Ans: We know that 

$\sqrt{\text{1+sin x}}\text{=cos}\frac{\text{x}}{\text{2}}\text{+sin}\frac{\text{x}}{\text{2}} $ 

$ \sqrt{\text{1-sin x}}\text{=cos}\frac{\text{x}}{\text{2}}\text{-sin}\frac{\text{x}}{\text{2}} $ 

Consider

$\text{co}{{\text{t}}^{\text{-1}}}\left( \frac{\sqrt{\text{1+sin x}}\text{+}\sqrt{\text{1-sin x}}}{\sqrt{\text{1+sin x}}\text{-}\sqrt{\text{1-sin x}}} \right) $ 

$ \text{=co}{{\text{t}}^{\text{-1}}}\left( \frac{\left( \text{cos}\frac{\text{x}}{\text{2}}\text{+sin}\frac{\text{x}}{\text{2}} \right)\text{+}\left( \text{cos}\frac{\text{x}}{\text{2}}\text{-sin}\frac{\text{x}}{\text{2}} \right)}{\left( \text{cos}\frac{\text{x}}{\text{2}}\text{+sin}\frac{\text{x}}{\text{2}} \right)\text{-}\left( \text{cos}\frac{\text{x}}{\text{2}}\text{-sin}\frac{\text{x}}{\text{2}} \right)} \right) $ 

$ \text{co}{{\text{t}}^{\text{-1}}}\left( \text{cot}\dfrac{\text{x}}{\text{2}} \right) $ 

$ \text{=}\dfrac{\text{x}}{\text{2}} $ 

It is proved that \[\text{co}{{\text{t}}^{\text{-1}}}\left( \dfrac{\sqrt{\text{1+sin x}}\text{+}\sqrt{\text{1-sin x}}}{\sqrt{\text{1+sin x}}\text{-}\sqrt{\text{1-sin x}}} \right)\text{=}\dfrac{\text{x}}{\text{2}}\]

10. Using inverse trigonometric identities, prove that ${{\tan }^{-1}}\left( \dfrac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}} \right)=\dfrac{\pi }{4}-\dfrac{1}{2}{{\cos }^{-1}}x,-\dfrac{1}{\sqrt{2}}\le x\le 1$

Ans: Assuming

$\text{x=cos2 }\!\!\beta\!\!\text{ } $ 

$ \sqrt{\text{1+x}} $ 

$ \text{=}\sqrt{\text{1+cos2 }\!\!\beta\!\!\text{ }} $ 

$ \text{=}\sqrt{\text{2}}\text{cos }\!\!\beta\!\!\text{ } $ 

$ \sqrt{\text{1-x}} $ 

$ \text{=}\sqrt{\text{1-cos2 }\!\!\beta\!\!\text{ }} $ 

$ \text{=}\sqrt{\text{2}}\text{sin }\!\!\beta\!\!\text{ } $

Consider

$\text{ta}{{\text{n}}^{\text{-1}}}\left( \dfrac{\sqrt{\text{1+x}}\text{-}\sqrt{\text{1-x}}}{\sqrt{\text{1+x}}\text{+}\sqrt{\text{1-x}}} \right) $ 

$\text{=ta}{{\text{n}}^{\text{-1}}}\left( \dfrac{\sqrt{\text{2}}\text{cos }\!\!\beta\!\!\text{ -}\sqrt{\text{2}}\text{sin }\!\!\beta\!\!\text{ }}{\sqrt{\text{2}}\text{cos }\!\!\beta\!\!\text{ +}\sqrt{\text{2}}\text{sin }\!\!\beta\!\!\text{ }} \right) $ 

$ \text{=ta}{{\text{n}}^{\text{-1}}}\left( \dfrac{\text{1-tan }\!\!\beta\!\!\text{ }}{\text{1+tan }\!\!\beta\!\!\text{ }} \right) $ 

$ \text{=ta}{{\text{n}}^{\text{-1}}}\left( \text{tan}\left( \dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}\text{- }\!\!\beta\!\!\text{ } \right) \right) $ 

$\text{=}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}\text{- }\!\!\beta\!\!\text{ } $ 

$ \text{=}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}\text{-}\dfrac{\text{1}}{\text{2}}\text{co}{{\text{s}}^{\text{-1}}}\text{x} $ 

It is proved that \[\text{ta}{{\text{n}}^{\text{-1}}}\left( \dfrac{\sqrt{\text{1+x}}\text{-}\sqrt{\text{1-x}}}{\sqrt{\text{1+x}}\text{+}\sqrt{\text{1-x}}} \right)\text{=}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}\text{-}\dfrac{\text{1}}{\text{2}}\text{co}{{\text{s}}^{\text{-1}}}\text{x}\]

11. For what value of $x$ does the equation $\text{2ta}{{\text{n}}^{\text{-1}}}\left( \text{cos x} \right)\text{=ta}{{\text{n}}^{\text{-1}}}\left( \text{2cosecx} \right)$ satisfy?

Ans: Consider \[\text{2ta}{{\text{n}}^{\text{-1}}}\left( \text{cos x} \right)\text{=ta}{{\text{n}}^{\text{-1}}}\left( \text{2cosecx} \right)\]

$ \text{ta}{{\text{n}}^{\text{-1}}}\left( \dfrac{\text{2cos x}}{\text{1-co}{{\text{s}}^{\text{2}}}\text{x}} \right)\text{=ta}{{\text{n}}^{\text{-1}}}\left( \text{2cosecx} \right) $ 

$ \dfrac{\text{2cos x}}{\text{1-co}{{\text{s}}^{\text{2}}}\text{x}}\text{=2cosecx} $ 

$ \text{sin xcos x=si}{{\text{n}}^{\text{2}}}\text{x} $ 

$ \text{sin x}\left( \text{cos x-sin x} \right)\text{=0} $ 

$ \text{sin x=0,cos x-sin x=0} $ 

$ \text{x=0,}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}} $ 

Therefore, \[\text{2ta}{{\text{n}}^{\text{-1}}}\left( \text{cos x} \right)\text{=ta}{{\text{n}}^{\text{-1}}}\left( \text{2cosecx} \right)\] is satisfied for \[\text{x=0,}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}\]

12. For what value of $x$ does the equation $\text{ta}{{\text{n}}^{\text{-1}}}\dfrac{\text{1-x}}{\text{1+x}}\text{=}\dfrac{\text{1}}{\text{2}}\text{ta}{{\text{n}}^{\text{-1}}}\text{x,}\left( \text{x > 0} \right)$ satisfy?

Ans: Consider \[\text{ta}{{\text{n}}^{\text{-1}}}\dfrac{\text{1-x}}{\text{1+x}}\text{=}\dfrac{\text{1}}{\text{2}}\text{ta}{{\text{n}}^{\text{-1}}}\text{x}\]

$ \text{ta}{{\text{n}}^{\text{-1}}}\left( \text{tan}\left( \dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}\text{-ta}{{\text{n}}^{\text{-1}}}\text{x} \right) \right)\text{=}\dfrac{\text{1}}{\text{2}}\text{ta}{{\text{n}}^{\text{-1}}}\text{x} $ 

$ \dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}\text{-ta}{{\text{n}}^{\text{-1}}}\text{x=}\dfrac{\text{1}}{\text{2}}\text{ta}{{\text{n}}^{\text{-1}}}\text{x} $ 

$ \dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}\text{=}\dfrac{\text{3}}{\text{2}}\text{ta}{{\text{n}}^{\text{-1}}}\text{x} $ 

$ \text{ta}{{\text{n}}^{\text{-1}}}\text{x=}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}} $ 

$ \text{x=}\dfrac{\text{1}}{\sqrt{\text{3}}} $ 

Therefore, \[\text{ta}{{\text{n}}^{\text{-1}}}\dfrac{\text{1-x}}{\text{1+x}}\text{=}\dfrac{\text{1}}{\text{2}}\text{ta}{{\text{n}}^{\text{-1}}}\text{x}\] is satisfied for \[\text{x=}\dfrac{\text{1}}{\sqrt{\text{3}}}\]

13. The expression $\text{sin}\left( \text{ta}{{\text{n}}^{\text{-1}}}\text{x} \right)\text{,}\left| \text{x} \right| < \text{1}$ is equal to

(A) $\dfrac{\text{x}}{\sqrt{\text{1-}{{\text{x}}^{\text{2}}}}}$

(B) $\dfrac{\text{1}}{\sqrt{\text{1-}{{\text{x}}^{\text{2}}}}}$

(C) $\dfrac{\text{1}}{\sqrt{\text{1+}{{\text{x}}^{\text{2}}}}}$

(D) $\dfrac{\text{x}}{\sqrt{\text{1+}{{\text{x}}^{\text{2}}}}}$

Ans: Assuming 

$ \text{x=tan }\!\!\beta\!\!\text{ } $ 

$ \text{ }\!\!\beta\!\!\text{ =ta}{{\text{n}}^{\text{-1}}}\text{x} $ 

$ \text{sin}\left( \text{ta}{{\text{n}}^{\text{-1}}}\text{x} \right) $ 

$ \text{=sin }\!\!\beta\!\!\text{ } $ 

$ \text{=}\dfrac{\text{x}}{\sqrt{\text{1+}{{\text{x}}^{\text{2}}}}} $ 

Hence, \[\text{sin}\left( \text{ta}{{\text{n}}^{\text{-1}}}\text{x} \right)\text{=}\dfrac{\text{x}}{\sqrt{\text{1+}{{\text{x}}^{\text{2}}}}}\]

Therefore, the correct option is option D

14. For what value of $x$ does the equation $\text{si}{{\text{n}}^{\text{-1}}}\left( \text{1-x} \right)\text{-2si}{{\text{n}}^{\text{-1}}}\text{x=}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}$ satisfy?

(A) $\text{0,}\dfrac{\text{1}}{\text{2}}$

(B) \[\text{1,}\dfrac{\text{1}}{\text{2}}\]

(C) \[0\]

(D) \[\dfrac{\text{1}}{\text{2}}\]

Ans: Consider

$ \text{si}{{\text{n}}^{\text{-1}}}\left( \text{1-x} \right)\text{-2si}{{\text{n}}^{\text{-1}}}\text{x=}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}} $ 

$\text{-2si}{{\text{n}}^{\text{-1}}}\text{x=}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}\text{-si}{{\text{n}}^{\text{-1}}}\left( \text{1-x} \right) $ 

\[\text{-2si}{{\text{n}}^{\text{-1}}}\text{x=co}{{\text{s}}^{\text{-1}}}\left( \text{1-x} \right)\] -----(1)

Assuming 

$ \text{co}{{\text{s}}^{\text{-1}}}\left( \text{1-x} \right)\text{= }\!\!\beta\!\!\text{ } $ 

$ \text{cos }\!\!\beta\!\!\text{ =1-x} $ 

$ \text{sin }\!\!\beta\!\!\text{ =}\sqrt{\text{1-}{{\left( \text{1-x} \right)}^{\text{2}}}} $ 

$ \text{sin }\!\!\beta\!\!\text{ =}\sqrt{\text{2x-}{{\text{x}}^{\text{2}}}} $ 

$ \text{ }\!\!\beta\!\!\text{ =si}{{\text{n}}^{\text{-1}}}\sqrt{\text{2x-}{{\text{x}}^{\text{2}}}} $ 

\[\text{co}{{\text{s}}^{\text{-1}}}\text{1-x=si}{{\text{n}}^{\text{-1}}}\sqrt{\text{2x-}{{\text{x}}^{\text{2}}}}\] -----(2)

Substituting Equation (2) in (1)

$\text{-2si}{{\text{n}}^{\text{-1}}}\text{x=si}{{\text{n}}^{\text{-1}}}\sqrt{\text{2x-}{{\text{x}}^{\text{2}}}} $ 

$ \text{si}{{\text{n}}^{\text{-1}}}\left( \text{-2x}\sqrt{\text{1-}{{\text{x}}^{\text{2}}}} \right)\text{=si}{{\text{n}}^{\text{-1}}}\sqrt{\text{2x-}{{\text{x}}^{\text{2}}}} $ 

$\text{-2x}\sqrt{\text{1-}{{\text{x}}^{\text{2}}}}\text{=}\sqrt{\text{2x-}{{\text{x}}^{\text{2}}}} $ 

$\text{4}{{\text{x}}^{\text{2}}}\text{-4}{{\text{x}}^{\text{4}}}\text{=2x-}{{\text{x}}^{\text{2}}} $ 

$ \text{4}{{\text{x}}^{\text{4}}}\text{-5}{{\text{x}}^{\text{2}}}\text{+2x=0} $ 

$ \text{x=0,}\dfrac{\text{1}}{\text{2}}\text{,}\dfrac{\text{-1 }\!\!\pm\!\!\text{ }\sqrt{\text{17}}}{\text{4}} $ 

But considering the given options and also when \[\text{x=}\dfrac{\text{1}}{\text{2}}\], the equation \[\text{si}{{\text{n}}^{\text{-1}}}\left( \text{1-x} \right)\text{-2si}{{\text{n}}^{\text{-1}}}\text{x=}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}\] doesn’t satisfy

Thus, $\text{x=0}$ is the only solution.

Hence the correct option is C

Conclusion

Miscellaneous Exercise in Class 12 Maths Chapter 2 is crucial for understanding various concepts thoroughly. It covers diverse problems that require the application of multiple formulas and techniques. It's important to focus on understanding the underlying principles behind each question rather than just memorizing solutions. Remember to understand the theory behind each concept, practice regularly, and refer to solved examples to master this exercise effectively.

Class 12 Maths Chapter 2: Exercises Breakdown

CBSE Class 12 Maths Chapter 2 Other Study Materials

Chapter-Specific NCERT Solutions for Class 12 Maths

Given below are the chapter-wise NCERT Solutions for Class 12 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.

Additional Study Materials for Class 12 Maths