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

Miscellaneous Solutions

1. Evaluate $$\text{co}{\text{s}}^{\text{-1}}\left(\text{cos}{\displaystyle \frac{\text{13 }\pi }{\text{6}}}\right)$$ 

Ans: Consider $\text{cos}{\displaystyle \frac{\text{13 }\pi }{\text{6}}}$

$\text{cos}{\displaystyle \frac{\text{13 }\pi }{\text{6}}}$ 

$\text{=cos}\left(\text{2 }\pi \text{ +}{\displaystyle \frac{\pi }{\text{6}}}\right)$ 

$\text{=cos}\left({\displaystyle \frac{\pi }{\text{6}}}\right)$ 

Further solving,

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

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

$\text{=}{\displaystyle \frac{\pi }{\text{6}}}$ 

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

2. Evaluate $$\text{ta}{\text{n}}^{\text{-1}}\left(\text{tan}{\displaystyle \frac{\text{7 }\pi }{\text{6}}}\right)$$ 

Ans: Consider $\text{tan}{\displaystyle \frac{\text{7 }\pi }{\text{6}}}$

$\text{tan}{\displaystyle \frac{\text{7 }\pi }{\text{6}}}$

$\text{=tan}\left(\pi \text{ +}{\displaystyle \frac{\pi }{\text{6}}}\right)$

$\text{=tan}\left({\displaystyle \frac{\pi }{\text{6}}}\right)$

Further solving,

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

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

$\text{=}{\displaystyle \frac{\pi }{\text{6}}}$

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

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

Ans: Assuming $$\text{2si}{\text{n}}^{\text{-1}}{\displaystyle \frac{\text{3}}{\text{5}}}\text{= }\alpha$$ ----(1)

$\text{sin}{\displaystyle \frac{\alpha }{\text{2}}}\text{=}{\displaystyle \frac{\text{3}}{\text{5}}}$ 

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

 $\text{cos}{\displaystyle \frac{\alpha }{\text{2}}}\text{=}{\displaystyle \frac{\text{4}}{\text{5}}}$ 

Hence,

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

$\text{sin }\alpha \text{ =}{\displaystyle \frac{\text{24}}{\text{25}}}$ 

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

$\text{cos }\alpha \text{ =}{\displaystyle \frac{\text{7}}{\text{25}}}$ 

Therefore, 

$$\text{tan }\alpha \text{ =}{\displaystyle \frac{\text{24}}{\text{7}}}$$

$$\alpha \text{ =ta}{\text{n}}^{\text{-1}}\left({\displaystyle \frac{\text{24}}{\text{7}}}\right)$$

From Equation (1), it is proved that 

$$\text{2si}{\text{n}}^{\text{-1}}{\displaystyle \frac{\text{3}}{\text{5}}}\text{=ta}{\text{n}}^{\text{-1}}\left({\displaystyle \frac{\text{24}}{\text{7}}}\right)$$

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

Ans: Assuming $$\text{si}{\text{n}}^{\text{-1}}{\displaystyle \frac{\text{8}}{\text{17}}}\text{= }\alpha$$ 

$\text{sin }\alpha \text{ =}{\displaystyle \frac{\text{8}}{\text{17}}}$ 

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

$\text{cos }\alpha \text{ =}{\displaystyle \frac{\text{15}}{\text{17}}}$ 

Hence,

Therefore, 

$\text{tan }\alpha \text{ =}{\displaystyle \frac{\text{8}}{\text{15}}}$ 

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

$$\text{si}{\text{n}}^{\text{-1}}\left({\displaystyle \frac{\text{8}}{\text{17}}}\right)\text{=ta}{\text{n}}^{\text{-1}}\left({\displaystyle \frac{\text{8}}{\text{15}}}\right)$$ ----(1) 

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

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

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

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

Therefore, 

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

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

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

Consider 

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

From Equations (1) and (2),

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

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

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

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

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

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

Ans: Assuming $$\text{co}{\text{s}}^{\text{-1}}{\displaystyle \frac{\text{4}}{\text{5}}}\text{= }\alpha$$ 

$\text{cos }\alpha \text{ =}{\displaystyle \frac{\text{4}}{\text{5}}}$ 

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

$\text{sin }\alpha \text{ =}{\displaystyle \frac{\text{3}}{\text{5}}}$ 

Therefore, 

$\text{tan }\alpha \text{ =}{\displaystyle \frac{\text{3}}{\text{4}}}$ 

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

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

Assuming that $$\text{co}{\text{s}}^{\text{-1}}{\displaystyle \frac{\text{12}}{\text{13}}}\text{= }\beta$$

$\text{cos }\beta \text{ =}{\displaystyle \frac{\text{12}}{\text{13}}}$ 

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

 $\text{sin }\beta \text{ =}{\displaystyle \frac{\text{5}}{\text{13}}}$ 

Therefore, 

$\text{tan }\beta \text{ =}{\displaystyle \frac{\text{5}}{\text{12}}}$ 

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

$$\text{co}{\text{s}}^{\text{-1}}\left({\displaystyle \frac{\text{12}}{\text{13}}}\right)\text{=ta}{\text{n}}^{\text{-1}}\left({\displaystyle \frac{\text{5}}{\text{12}}}\right)$$ ----(2) 

Consider 

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

From Equations (1) and (2),

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

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

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

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

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

Hence, it is proved that $$\text{co}{\text{s}}^{\text{-1}}{\displaystyle \frac{\text{4}}{\text{5}}}\text{+co}{\text{s}}^{\text{-1}}{\displaystyle \frac{\text{12}}{\text{13}}}\text{=co}{\text{s}}^{\text{-1}}{\displaystyle \frac{\text{33}}{\text{65}}}$$

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

Ans: Assuming $$\text{co}{\text{s}}^{\text{-1}}{\displaystyle \frac{\text{12}}{\text{13}}}\text{= }\alpha$$ 

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

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

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

Therefore, 

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

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

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

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

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

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

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

Therefore, 

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

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

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

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

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

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

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

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

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 $$\text{si}{\text{n}}^{\text{-1}}\frac{\text{5}}{\text{13}}\text{= }\alpha$$ 

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

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

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

Therefore, 

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

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

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

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

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

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

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

Therefore, 

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

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

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

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

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

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

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

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 ${\tan }^{-1}\sqrt{x}=\frac{1}{2}{\cos }^{-1}\left(\frac{1-x}{1+x}\right),x\in [0,1]$

Ans: Assuming $$\text{x=ta}{\text{n}}^{\text{2}}\alpha$$ 

Consider

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

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

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

$\text{= }\alpha$ 

$$\text{ta}{\text{n}}^{\text{-1}}\sqrt{\text{x}}\text{= }\alpha$$----(1)

Consider 

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

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

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

$\text{= }\alpha$ 

$$\frac{\text{1}}{\text{2}}\text{co}{\text{s}}^{\text{-1}}\left(\frac{\text{1-x}}{\text{1+x}}\right)\text{= }\alpha$$ -----(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 ${\cot }^{-1}\left(\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right)=\frac{x}{2},x\in (0,\frac{\pi }{4})$

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

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

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

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

$\text{=}{\displaystyle \frac{\text{x}}{\text{2}}}$ 

It is proved that $$\text{co}{\text{t}}^{\text{-1}}\left({\displaystyle \frac{\sqrt{\text{1+sin x}}\text{+}\sqrt{\text{1-sin x}}}{\sqrt{\text{1+sin x}}\text{-}\sqrt{\text{1-sin x}}}}\right)\text{=}{\displaystyle \frac{\text{x}}{\text{2}}}$$

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

Ans: Assuming

$\text{x=cos2 }\beta$ 

$\sqrt{\text{1+x}}$ 

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

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

$\sqrt{\text{1-x}}$ 

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

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

Consider

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

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

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

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

$\text{=}{\displaystyle \frac{\pi }{\text{4}}}\text{- }\beta$ 

$\text{=}{\displaystyle \frac{\pi }{\text{4}}}\text{-}{\displaystyle \frac{\text{1}}{\text{2}}}\text{co}{\text{s}}^{\text{-1}}\text{x}$ 

It is proved that $$\text{ta}{\text{n}}^{\text{-1}}\left({\displaystyle \frac{\sqrt{\text{1+x}}\text{-}\sqrt{\text{1-x}}}{\sqrt{\text{1+x}}\text{+}\sqrt{\text{1-x}}}}\right)\text{=}{\displaystyle \frac{\pi }{\text{4}}}\text{-}{\displaystyle \frac{\text{1}}{\text{2}}}\text{co}{\text{s}}^{\text{-1}}\text{x}$$

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

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

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

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

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

$\text{sin x=0,cos x-sin x=0}$ 

$\text{x=0,}{\displaystyle \frac{\pi }{\text{4}}}$ 

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 $$\text{x=0,}{\displaystyle \frac{\pi }{\text{4}}}$$

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

Ans: Consider $$\text{ta}{\text{n}}^{\text{-1}}{\displaystyle \frac{\text{1-x}}{\text{1+x}}}\text{=}{\displaystyle \frac{\text{1}}{\text{2}}}\text{ta}{\text{n}}^{\text{-1}}\text{x}$$

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

${\displaystyle \frac{\pi }{\text{4}}}\text{-ta}{\text{n}}^{\text{-1}}\text{x=}{\displaystyle \frac{\text{1}}{\text{2}}}\text{ta}{\text{n}}^{\text{-1}}\text{x}$ 

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

$\text{ta}{\text{n}}^{\text{-1}}\text{x=}{\displaystyle \frac{\pi }{\text{6}}}$ 

$\text{x=}{\displaystyle \frac{\text{1}}{\sqrt{\text{3}}}}$ 

Therefore, $$\text{ta}{\text{n}}^{\text{-1}}{\displaystyle \frac{\text{1-x}}{\text{1+x}}}\text{=}{\displaystyle \frac{\text{1}}{\text{2}}}\text{ta}{\text{n}}^{\text{-1}}\text{x}$$ is satisfied for $$\text{x=}{\displaystyle \frac{\text{1}}{\sqrt{\text{3}}}}$$

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) ${\displaystyle \frac{\text{x}}{\sqrt{\text{1-}{\text{x}}^{\text{2}}}}}$

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

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

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

Ans: Assuming 

$\text{x=tan }\beta$ 

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

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

$\text{=sin }\beta$ 

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

Hence, $$\text{sin}\left(\text{ta}{\text{n}}^{\text{-1}}\text{x}\right)\text{=}{\displaystyle \frac{\text{x}}{\sqrt{\text{1+}{\text{x}}^{\text{2}}}}}$$

Therefore, the correct option is option D

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

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

(B) $$\text{1,}{\displaystyle \frac{\text{1}}{\text{2}}}$$

(C) $$0$$

(D) $${\displaystyle \frac{\text{1}}{\text{2}}}$$

Ans: Consider

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

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

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

Assuming 

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

$\text{cos }\beta \text{ =1-x}$ 

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

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

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

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

Substituting Equation (2) in (1)

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

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

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

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

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

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

But considering the given options and also when $$\text{x=}{\displaystyle \frac{\text{1}}{\text{2}}}$$, the equation $$\text{si}{\text{n}}^{\text{-1}}\left(\text{1-x}\right)\text{-2si}{\text{n}}^{\text{-1}}\text{x=}{\displaystyle \frac{\pi }{\text{2}}}$$ 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