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NCERT Solutions for Class 12 Maths Miscellaneous Exercise Chapter 2- Inverse Trigonometric Functions

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NCERT Solutions for Class 12 Maths Chapter 2 Miscellaneous Exercise - Free PDF Download

NCERT Solutions for Class 12 Maths Chapter 2 Inverse Trigonometric Functions includes solutions to all Miscellaneous Exercise problems. Inverse Trigonometric Functions Class 12 NCERT Solutions Miscellaneous Exercises are based on the ideas presented in Maths Chapter 2. This activity is crucial for both the CBSE Board examinations and competitive tests. To perform well on the board exam, download the NCERT Solutions for Class 12 Maths Inverse Trigonometric Functions Miscellaneous Exercise in PDF format and practice them offline.

Table of Content
1. NCERT Solutions for Class 12 Maths Chapter 2 Miscellaneous Exercise - Free PDF Download
2. Access NCERT Class 12 Maths Chapter 2 Inverse Trigonometric Functions Miscellaneous Exercise
3. Class 12 Maths Chapter 2: Exercises Breakdown
4. CBSE Class 12 Maths Chapter 2 Other Study Materials
5. Chapter-Specific NCERT Solutions for Class 12 Maths
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Access NCERT Class 12 Maths Chapter 2 Inverse Trigonometric Functions Miscellaneous Exercise

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

 Exercise Number of Questions Exercise 2.1 14 Questions and Solutions Exercise 2.2 15 Questions and Solutions

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.

FAQs on NCERT Solutions for Class 12 Maths Miscellaneous Exercise Chapter 2- Inverse Trigonometric Functions

1. Can you give an example of how to use the NCERT solutions for a typical problem?

While I cannot provide specific solutions due to copyright, I can explain the approach using a general example. Suppose a question asks for the value of $\sin^{-1}$ (1/2). Here, you'd use the concept of the inverse sine function to find the angle whose sine value is 1/2. Referring to the unit circle in your NCERT book, you'll find that the angle whose sine is 1/2 is π/6. The solutions would guide you through this process.

2. How can I solve problems involving functions like $\cos^{-1}$ (-√3/2)?

Similar to the previous question, you'll use the inverse cosine function to find the angle whose cosine is -√3/2. The solutions will provide guidance, but remember that cosine repeats every 2π. This means there might be multiple solutions.The principal value range (0 to π) for $\cos^{-1}$ (-√3/2) is 5π/6, but depending on the context, other answers like 7π/6 might also be valid.

3. Can the NCERT solutions help me solve for $\tan^{-1}$ (1)?

Yes, the solutions can definitely help. The inverse tangent function $\tan^{-1}$ gives the angle whose tangent is the given value. The solutions would guide you towards finding the answer, which is π/4 for $\tan^{-1}$ (1).

4. What approach should I take for problems involving trigonometric identities?

The NCERT Solutions often use trigonometric identities to simplify expressions before applying inverse trigonometric functions. It's important to have a strong understanding of these identities. Revisit the relevant sections in your NCERT book to solidify your grasp on these identities before tackling such problems in the solutions.

5. Where can I find solutions to other problems in the miscellaneous exercise?

The NCERT textbook itself provides solutions to some problems within the miscellaneous exercise section. Additionally, Vedantu offers various online resources for comprehensive solutions. You can find them through a web search using terms like "NCERT Solutions Class 12 Maths Chapter 2 Miscellaneous Exercise".

By understanding these approaches and utilizing the NCERT solutions along with other resources, you'll be well-equipped to conquer the challenges presented in the miscellaneous exercise section of Chapter 2 on Inverse Trigonometric Functions.