Tag: Miscellaneous Exercise on chapter 2 solution maths class 12th ncert

1. ${\cos }^{-1}(\cos (13\pi \mathrm{/}6))$

We know that ${\cos }^{-1}(\cos \theta )=\theta$ if $\theta \in [0,\pi ]$.

$$13\pi \mathrm{/}6=2\pi +\pi \mathrm{/}6⟹\cos (13\pi \mathrm{/}6)=\cos (\pi \mathrm{/}6)$$

Thus,

$${\cos }^{-1}(\cos (13\pi \mathrm{/}6))=\pi \mathrm{/}6$$

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2. ${\tan }^{-1}(\tan (7\pi \mathrm{/}6))$

${\tan }^{-1}(\tan \theta )=\theta$ if $\theta \in (-\pi \mathrm{/}2,\pi \mathrm{/}2)$.

$$7\pi \mathrm{/}6=\pi +\pi \mathrm{/}6⟹\tan (7\pi \mathrm{/}6)=\tan (\pi \mathrm{/}6)$$

So,

$${\tan }^{-1}(\tan (7\pi \mathrm{/}6))=\pi \mathrm{/}6-\pi =-5\pi \mathrm{/}6$$

Since range is $(-\pi \mathrm{/}2,\pi \mathrm{/}2)$, principal value is $\pi \mathrm{/}6-\pi =-\pi \mathrm{/}6$

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3. Prove $2{\sin }^{-1}\left(\frac{3}{5}\right)={\tan }^{-1}\left(\frac{24}{7}\right)$

Let ${\sin }^{-1}\frac{3}{5}=\theta$, so $\sin \theta =\frac{3}{5}$.
Then, $\cos \theta =\frac{4}{5}$

$$\tan (2\theta )=\frac{2\tan \theta }{1-{\tan }^2\theta }=\frac{2(\frac{3}{4})}{1-(\frac{3}{4}{)}^2}=\frac{24}{7}$$

Hence,

$$2{\sin }^{-1}\left(\frac{3}{5}\right)={\tan }^{-1}\left(\frac{24}{7}\right)$$

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4. Prove ${\sin }^{-1}\frac{8}{17}+{\sin }^{-1}\frac{3}{5}={\tan }^{-1}\frac{77}{36}$

Let ${\sin }^{-1}\frac{8}{17}=\alpha$ and ${\sin }^{-1}\frac{3}{5}=\beta$

$$\tan \alpha =\frac{8}{15},\tan \beta =\frac{3}{4}$$

Thus,

$$\tan (\alpha +\beta )=\frac{8\mathrm{/}15+3\mathrm{/}4}{1-(8\mathrm{/}15)(3\mathrm{/}4)}=\frac{77}{36}$$

Hence proved.

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5. Prove ${\cos }^{-1}\frac{4}{5}+{\cos }^{-1}\frac{12}{13}={\cos }^{-1}\frac{33}{65}$

Let $A={\cos }^{-1}\frac{4}{5},B={\cos }^{-1}\frac{12}{13}$.

$$\cos (A+B)=\cos A\cos B-\sin A\sin B$$
$$=\frac{4}{5}\cdot \frac{12}{13}-\frac{3}{5}\cdot \frac{5}{13}=\frac{33}{65}$$

Hence proved.

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6. ${\cos }^{-1}\frac{12}{13}+{\sin }^{-1}\frac{3}{5}={\sin }^{-1}\frac{56}{65}$

Let $A={\cos }^{-1}\frac{12}{13},B={\sin }^{-1}\frac{3}{5}$
Then $\sin A=\frac{5}{13},\cos B=\frac{4}{5}$

$$\sin (A+B)=\sin A\cos B+\cos A\sin B=\frac{5}{13}\cdot \frac{4}{5}+\frac{12}{13}\cdot \frac{3}{5}=\frac{56}{65}$$

Hence proved.

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7. ${\tan }^{-1}\frac{63}{16}={\sin }^{-1}\frac{5}{13}+{\cos }^{-1}\frac{3}{5}$

Let ${\sin }^{-1}\frac{5}{13}=A$, ${\cos }^{-1}\frac{3}{5}=B$$$\tan (A+B)=\frac{5\mathrm{/}12+4\mathrm{/}3}{1-\frac{5}{12}\cdot \frac{4}{3}}=\frac{63}{16}$$

Hence proved.

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8. Prove ${\tan }^{-1}\sqrt{\frac{1-x}{1+x}}=\frac{1}{2}{\cos }^{-1}x,x\in (0,1)$

Put $x=\cos 2\theta$, then

$$\sqrt{\frac{1-x}{1+x}}=\sqrt{\frac{1-\cos 2\theta }{1+\cos 2\theta }}=\tan \theta$$

Thus,

$${\tan }^{-1}\sqrt{\frac{1-x}{1+x}}=\theta =\frac{1}{2}{\cos }^{-1}x$$

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9. Prove ${\cot }^{-1}\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}=\frac{x}{2}$Let $t=\tan \frac{x}{2}$, then:

$$\sin x=\frac{2t}{1+t^2}$$

After rationalizing and simplifying, the expression equals ${\cot }^{-1}\left(\frac{1}{\tan (x\mathrm{/}2)}\right)=\frac{x}{2}$

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10. Prove ${\tan }^{-1}\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}=\frac{\pi }{4}-\frac{1}{2}{\cos }^{-1}x$

Let $x=\cos 2\theta$, then

$$\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}=\frac{\cos \theta -\sin \theta }{\cos \theta +\sin \theta }=\tan (\frac{\pi }{4}-\theta )$$

Hence,

$${\tan }^{-1}\text{(LHS)}=\frac{\pi }{4}-\theta =\frac{\pi }{4}-\frac{1}{2}{\cos }^{-1}x$$

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11. Solve $2{\tan }^{-1}(\cos x)={\tan }^{-1}(2\csc x)$

Let ${\tan }^{-1}(\cos x)=\theta$

Then $\tan (2\theta )=\frac{2\tan \theta }{1-{\tan }^2\theta }=\frac{2\cos x}{1-{\cos }^{2}x}=2\csc x$

Hence proven.

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12. Prove ${\tan }^{-1}\frac{1-x}{1+x}=\frac{\pi }{4}-{\tan }^{-1}x$

Use the tangent subtraction identity:

$$\tan (\frac{\pi }{4}-{\tan }^{-1}x)=\frac{1-x}{1+x}$$

Taking ${\tan }^{-1}$ both sides gives the result.

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13. $\sin ({\tan }^{-1}x)=\frac{x}{\sqrt{1+x^2}}$

Let ${\tan }^{-1}x=\theta \Rightarrow \tan \theta =x$

So a right-angle triangle gives $\sin \theta =\frac{x}{\sqrt{1+x^2}}$

Hence,

$$\sin ({\tan }^{-1}x)=\frac{x}{\sqrt{1+x^2}}$$

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14. If ${\sin }^{-1}(1-2x^2)=2{\sin }^{-1}x$

Use double angle formula:

$$\sin (2{\sin }^{-1}x)=2x\sqrt{1-x^2}$$

Equating $1-2x^2=2x\sqrt{1-x^2}$ and solving gives $x=0$ or $x=1$.