Chapter 2: Miscellaneous Exercise

Find the value of the following:(Class 12 ncert solution math chapter 2 Miscellaneous)

Question 1: Find the value of $\cos ^{-1}\left(\cos \frac{13 \pi}{6}\right)$ .

Solution: $\cos ^{-1}\left(\cos \frac{13 \pi}{6}\right) =\cos ^{-1}\left[\cos \left(2 \pi+\frac{\pi}{6}\right)\right]$

$=\cos ^{-1}\left[\cos \frac{\pi}{6}\right]$

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

Question 2: Find the value of $\tan ^{-1}\left(\tan \frac{7 \pi}{6}\right)$ .

Solution: $\left.\tan^{1}\left(\tan \frac{7 \pi}{6}\right) =\tan ^{-1}\left[\tan \left(2 \pi-\frac{5 \pi}{6}\right)\right]$

$=\tan ^{-1}\left[-\tan \left(\frac{5 \pi}{6}\right)\right]$

$=\tan ^{-1}\left[\tan \left(\pi-\frac{5 \pi}{6}\right)\right]$

$=\tan ^{-1}\left[\tan \frac{\pi}{6}\right]$

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

Question 3: Prove that $2 \sin ^{-1} \frac{3}{5}=\tan ^{-1} \frac{24}{7}$ .

Solution: Let $\sin ^{-1} \frac{3}{5}=x \Rightarrow \sin x=\frac{3}{5}$

Then,

$\cos x=\sqrt{1-\left(\frac{3}{5}\right)^{2}}=\frac{4}{5}$

Therefore,

Thus,

$\tan x =\frac{3}{4}$

$x =\tan ^{-1} \frac{3}{4}$

$\sin ^{-1} \frac{3}{5} =\tan ^{-1} \frac{3}{4}$

$LHS =2 \sin ^{-1} \frac{3}{5}$

$=2 \tan ^{-1} \frac{3}{4}$

$=\tan ^{-1}\left(\frac{2 \times \frac{3}{4}}{1-\left(\frac{3}{4}\right)^{2}}\right)$

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

$=R H S$

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

Solution: Let $\sin ^{-1} \frac{8}{17}=x \Rightarrow \sin x=\frac{8}{17}$

Then,

Therefore,

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

$=\sqrt{\frac{225}{289}}=\frac{15}{17}$

$\tan x =\frac{8}{15}$

$x =\tan ^{-1} \frac{8}{15}$

$\sin ^{-1} \frac{8}{17} =\tan ^{-1} \frac{8}{15}$

Now, let $\sin ^{-1} \frac{3}{5}=y \Rightarrow \sin y=\frac{3}{5}$

Then,

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

$=\sqrt{\frac{16}{25}}=\frac{4}{5}$

Therefore,

$\tan y =\frac{3}{4}$

$y =\tan ^{-1} \frac{3}{4}$

$\sin ^{-1} \frac{3}{5} =\tan ^{-1} \frac{3}{4}$

Thus, by using (1) and (2)

$LHS =\sin ^{-1} \frac{8}{17}+\sin ^{-1} \frac{3}{5}$

$=\tan ^{-1} \frac{8}{15}+\tan ^{-1} \frac{3}{4}$

$=\tan ^{-1}\left[\frac{\frac{8}{15}+\frac{3}{4}}{1-\frac{8}{15} \cdot \frac{3}{4}}\right]$

$=\tan ^{-1}\left[\frac{\frac{32+45}{60}}{60-24}\right]$

$=\tan ^{-1} \frac{77}{36}$

$=R H S$

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

Solution: Let $\cos ^{-1} \frac{4}{5}=x \Rightarrow \cos x=\frac{4}{5}$

Then, Therefore,

$\sin x=\sqrt{1-\left(\frac{4}{5}\right)^{2}}=\frac{3}{5}$

$\tan x =\frac{3}{4}$

$x =\tan ^{-1} \frac{3}{4}$

$\cos ^{-1} \frac{4}{5} =\tan ^{-1} \frac{3}{4}$ —(i)

Now, let $\cos ^{-1} \frac{12}{13}=y \Rightarrow \cos y=\frac{12}{13}$

Then,

Therefore,

$\sin y=\frac{5}{13}$

$\tan y =\frac{5}{12}$

$y =\tan ^{-1} \frac{5}{12}$

$\cos ^{-1} \frac{12}{13} =\tan ^{-1} \frac{5}{12}$ —(ii)

Thus, by using (1) and (2)

$\cos ^{-1} \frac{4}{5}+\cos ^{-1} \frac{12}{13} =\tan ^{-1} \frac{3}{4}+\tan ^{-1} \frac{5}{12}$

$=\tan ^{-1}\left[\frac{\frac{3}{4}+\frac{5}{12}}{1-\frac{3}{4} \cdot \frac{5}{12}}\right]$

$=\tan ^{-1}\left[\frac{56}{33}\right]$

Now, let $\cos ^{-1} \frac{33}{65}=z \Rightarrow \cos z=\frac{33}{65}$

Then,

$\sin z=\sqrt{1-\left(\frac{33}{65}\right)^{2}}=\frac{56}{65}$

Therefore,

$\tan z =\frac{33}{56}$

$z =\tan ^{-1} \frac{56}{33}$

$\cos ^{-1} \frac{33}{65} =\tan ^{-1} \frac{56}{33}$

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

Hence proved.

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

Solution: Let $\cos ^{-1} \frac{12}{13}=y \Rightarrow \cos y=\frac{12}{13}$

Then,

Therefore,

$\sin y=\sqrt{1-\left(\frac{12}{13}\right)^{2}}=\frac{5}{13}$

$\tan y =\frac{5}{12}$

$y =\tan ^{-1} \frac{5}{12}$

$\cos ^{-1} \frac{12}{13} =\tan ^{-1} \frac{5}{12}$

Now, let $\sin ^{-1} \frac{3}{5}=x \Rightarrow \sin x=\frac{3}{5}$

Then,

Therefore,

$\cos x=\sqrt{1-\left(\frac{3}{5}\right)^{2}}=\frac{4}{5}$

$\tan x =\frac{3}{4}$

$x =\tan ^{-1} \frac{3}{4}$

$\sin ^{-1} \frac{3}{5} =\tan ^{-1} \frac{3}{4}$

Now, let $\sin ^{-1} \frac{56}{65}=z \Rightarrow \sin z=\frac{56}{65}$

Then,

Therefore,

$\cos z=\sqrt{1-\left(\frac{56}{65}\right)^{2}}=\frac{33}{65}$

$\tan z=\frac{56}{33}$

$z=\tan ^{-1} \frac{56}{33}$

$\sin ^{-1} \frac{56}{65}=\tan ^{-1} \frac{56}{33}$

Thus, by using (1) and (2)

$LHS =\cos ^{-1} \frac{12}{13}+\sin ^{-1} \frac{3}{5}$

$=\tan ^{-1} \frac{5}{12}+\tan ^{-1} \frac{3}{4}$

$=\tan ^{-1}\left[\frac{\frac{5}{12}+\frac{3}{4}}{1-\frac{5}{12} \cdot \frac{3}{4}}\right]$

$\left.=\tan ^{-1}\left[\frac{\frac{20+36}{48-15}}{48}\right] \quad \text { [Using }(3)\right]$

$=\tan ^{-1}\left(\frac{56}{33}\right)$

$=\sin ^{-1} \frac{56}{65}$

$=R H S$

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

Solution: Let $\sin ^{-1} \frac{5}{13}=x \Rightarrow \sin x=\frac{5}{13}$

Then,

Therefore,

$\cos x=\sqrt{1-\left(\frac{5}{13}\right)^{2}}=\frac{12}{13}$

$\tan x =\frac{5}{12}$

$x =\tan ^{-1} \frac{5}{12}$

$\sin ^{-1} \frac{5}{13} =\tan ^{-1} \frac{5}{12}$

Now, let $\cos ^{-1} \frac{3}{5}=y \Rightarrow \cos y=\frac{3}{5}$

Then,

$\sin y=\sqrt{1-\left(\frac{3}{5}\right)^{2}}=\frac{4}{5}$

Therefore,

$\tan y =\frac{4}{3}$

$y =\tan ^{-1} \frac{4}{3}$

$\cos ^{-1} \frac{3}{5} =\tan ^{-1} \frac{4}{3}$

Thus, by using (1) and (2)

$RHS =\sin ^{-1} \frac{5}{12}+\cos ^{-1} \frac{3}{5}$

$=\tan ^{-1} \frac{5}{12}+\tan ^{-1} \frac{4}{3}$

$=\tan ^{-1}\left(\frac{\frac{5}{12}+\frac{4}{3}}{1-\frac{5}{12} \cdot \frac{4}{3}}\right)$

$=\tan ^{-1}\left(\frac{63}{16}\right)$

$=L H S$

Question 8: Prove that $\tan ^{-1} \frac{1}{5}+\tan ^{-1} \frac{1}{7}+\tan ^{-1} \frac{1}{3}+\tan ^{-1} \frac{1}{8}=\frac{\pi}{4}$

Solution:

$LHS =\tan ^{-1} \frac{1}{5}+\tan ^{-1} \frac{1}{7}+\tan ^{-1} \frac{1}{3}+\tan ^{-1} \frac{1}{8}$

$=\tan ^{-1}\left(\frac{\frac{1}{5}+\frac{1}{7}}{1-\frac{1}{5} \cdot \frac{1}{7}}\right)+\tan ^{-1}\left(\frac{\frac{1}{3}+\frac{1}{8}}{1-\frac{1}{3} \cdot \frac{1}{8}}\right)$

$=\tan ^{-1}\left(\frac{12}{34}\right)+\tan ^{-1}\left(\frac{11}{23}\right)$

$=\tan ^{-1}\left(\frac{6}{17}\right)+\tan ^{-1}\left(\frac{11}{23}\right)$

$=\tan ^{-1}\left(\frac{\frac{6}{17}+\frac{11}{23}}{1-\frac{6}{17} \cdot \frac{11}{23}}\right)$

$=\tan ^{-1}\left(\frac{325}{325}\right)$

$=\tan ^{-1}(1)$

$=\frac{\pi}{4}$

$=R H S$

Question 9: Prove that $\tan ^{-1} \sqrt{x}=\frac{1}{2} \cos ^{-1}\left(\frac{1-x}{1+x}\right), x \in[0,1]$ .

Solution: Let $x=\tan ^{2} \theta$

Then,

Therefore,

$\sqrt{x} =\tan \theta$

$\theta =\tan ^{-1} \sqrt{x}$

Thus,

$\left(\frac{1-x}{1+x}\right) =\frac{1-\tan ^{2} \theta}{1+\tan ^{2} \theta}$

$=\cos 2 \theta$

$RHS =\frac{1}{2} \cos ^{-1}\left(\frac{1-x}{1+x}\right)$

$=\frac{1}{2} \cos ^{-1}(\cos 2 \theta)$

$=\frac{1}{2} \times 2 \theta$

$=\theta$

$=\tan ^{-1} \sqrt{x}=L H S$

Question 10: Prove that $\cot ^{-1}\left(\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right)=\frac{x}{2}, x \in\left(0, \frac{\pi}{4}\right)$ .

Solution:

$\left(\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right) =\frac{(\sqrt{1+\sin x}+\sqrt{1-\sin x})^{2}}{(\sqrt{1+\sin x})^{2}-(\sqrt{1-\sin x})^{2}} \quad \text { (by rationalizing) }$

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

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

$=\frac{2 \cos ^{2} \frac{x}{2}}{2 \sin \frac{x}{2} \cos \frac{x}{2}}$

$=\cot \frac{x}{2}$

Thus,

$L H S =\cot ^{-1}\left(\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right)$

$=\cot ^{-1}\left(\cot \frac{x}{2}\right)$

$=\frac{x}{2}=R H S$

Question 11: Prove that $\tan ^{-1}\left(\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right)=\frac{\pi}{2}-\frac{1}{2} \cos ^{-1} x,-\frac{1}{\sqrt{2}} \leq x \leq 1$

Solution: Let $x=\cos 2 \theta \Rightarrow \theta=\frac{1}{2} \cos ^{-1} x$

Thus,

$L H S =\tan ^{-1}\left(\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right)$

$=\tan ^{-1}\left(\frac{\sqrt{1+\cos 2 \theta}-\sqrt{1-\cos 2 \theta}}{\sqrt{1+\cos 2 \theta}+\sqrt{1-\cos 2 \theta}}\right)$

$=\tan ^{-1}\left(\frac{\sqrt{2 \cos ^{2} \theta}-\sqrt{2 \sin ^{2} \theta}}{\sqrt{2 \cos ^{2} \theta}+\sqrt{2 \sin ^{2} \theta}}\right)$

$=\tan ^{-1}\left(\frac{\sqrt{2} \cos \theta-\sqrt{2} \sin \theta}{\sqrt{2} \cos \theta+\sqrt{2} \sin \theta}\right)$

$=\tan ^{-1}\left(\frac{\cos \theta-\sin \theta}{\cos \theta+\sin \theta}\right)$

$=\tan ^{-1}\left(\frac{1-\tan \theta}{1+\tan \theta}\right)$

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

$=\frac{\pi}{4}-\theta$

$=\frac{\pi}{4}-\frac{1}{2} \cos^{-1}x =R H S$

Question 12: Prove that $\frac{9 \pi}{8}-\frac{9}{4} \sin ^{-1} \frac{1}{3}=\frac{9}{4} \sin ^{-1} \frac{2 \sqrt{2}}{3}$

Solution: $LHS =\frac{9 \pi}{8}-\frac{9}{4} \sin ^{-1} \frac{1}{3}$

$=\frac{9}{4}\left(\frac{\pi}{2}-\sin ^{-1} \frac{1}{3}\right)$

$=\frac{9}{4}\left(\cos ^{-1} \frac{1}{3}\right)$

Now, let $\cos ^{-1} \frac{1}{3}=x \Rightarrow \cos x=\frac{1}{3}$

Therefore,

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

$=\frac{2 \sqrt{2}}{3}$

$x =\sin ^{-1} \frac{2 \sqrt{2}}{3}$

$\cos ^{-1} \frac{1}{3} =\sin ^{-1} \frac{2 \sqrt{2}}{3}$

Thus, by using (1) and (2)

$\frac{9 \pi}{8}-\frac{9}{4} \sin ^{-1} \frac{1}{3}=\frac{9}{4} \sin ^{-1} \frac{2 \sqrt{2}}{3}$

Hence proved.

Question 13: Solve $2 \tan ^{-1}(\cos x)=\tan ^{-1}(2 \operatorname{cosec} x)$ .

Solution: It is given that $2 \tan ^{-1}(\cos x)=\tan ^{-1}(2 \operatorname{cosec} x)$

Since, $2 \tan ^{-1}(x)=\tan ^{-1} \frac{2 x}{1-x^{2}}$

Hence, Therefore,

$\Rightarrow \tan ^{-1}\left(\frac{2 \cos x}{1-\cos ^{2} x}\right)=\tan ^{-1}(2 \operatorname{cosec} x)$

$\Rightarrow\left(\frac{2 \cos x}{1-\cos ^{2} x}\right)=(2 \operatorname{cosec} x)$

$\Rightarrow \frac{2 \cos x}{\sin ^{2} x}=\frac{2}{\sin x}$

$\Rightarrow \cos x=\sin x$

$\Rightarrow \tan x=1$

$\Rightarrow \tan x=\tan \frac{\pi}{4}$

$x=n \pi+\frac{\pi}{4} \text {, where } n \in Z \text {. }$

Question 14: Solve $\tan ^{-1} \frac{1-x}{1+x}=\frac{1}{2} \tan ^{-1} x,(x>0)$

Solution: Since ${ }^{\tan ^{-1} x-\tan ^{-1} y=\tan ^{-1} \frac{x-y}{1+x y}}$

Hence,

$\Rightarrow \tan ^{-1} \frac{1-x}{1+x}=\frac{1}{2} \tan ^{-1} x$

$\Rightarrow \tan ^{-1} 1-\tan ^{-1} x=\frac{1}{2} \tan ^{-1} x$

$\Rightarrow \frac{\pi}{4}=\frac{3}{2} \tan ^{-1} x$

$\Rightarrow \tan ^{-1} x=\frac{\pi}{6}$

$\Rightarrow x=\tan \frac{\pi}{6}$

$\Rightarrow x=\frac{1}{\sqrt{3}}$

Question 15: Solve $\sin \left(\tan ^{-1} x\right),|x|<1$ is equal to

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

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

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

Solution: Let $\tan y=x$

Therefore,

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

Now, let $\tan ^{-1} x=y$

Therefore,

Hence,

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

Thus,

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

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

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

Thus, the correct option is $\mathrm{D}$ .

Question 16: Solve: $\sin ^{-1}(1-x)-2 \sin ^{-1} x=\frac{\pi}{2}$ , then $x$ is equal to

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

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

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

Solution: It is given that $\sin ^{-1}(1-x)-2 \sin ^{-1} x=\frac{\pi}{2}$

$\Rightarrow \sin ^{-1}(1-x)-2 \sin ^{-1} x=\frac{\pi}{2}$

$\Rightarrow-2 \sin ^{-1} x=\frac{\pi}{2}-\sin ^{-1}(1-x)$

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

Let $\sin ^{-1} x=y \Rightarrow \sin y=x$

Hence,

$\cos y =\sqrt{1-x^{2}}$

$y =\cos ^{-1}\left(\sqrt{1-x^{2}}\right)$

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

From equation (1), we have

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

Put $x=\sin y$

Therefore,

$\Rightarrow-2 \cos ^{-1} \sqrt{1-\sin ^{2} y}=\cos ^{-1}(1-\sin y)$

$\Rightarrow-2 \cos ^{-1}(\cos y)=\cos ^{-1}(1-\sin y)$

$\Rightarrow-2 y=\cos ^{-1}(1-\sin y)$

$\Rightarrow 1-\sin y=\cos (-2 y)$

$\Rightarrow 1-\sin y=\cos 2 y$

$\Rightarrow 1-\sin y=1-2 \sin ^{2} y$

$\Rightarrow 2 \sin 2 y-\sin y=0$

$\Rightarrow \sin y(2 \sin y-1)=0$

$\Rightarrow \sin y=0, \frac{1}{2}$

$x=0, \frac{1}{2}$

When $x=\frac{1}{2}$ , it does not satisfy the equation.

Hence, $x=0$ is the only solution

Thus, the correct option is $\mathrm{C}$ .

Question 17: Solve $\tan ^{-1}\left(\frac{x}{y}\right)-\tan ^{-1} \frac{x-y}{x+y}$ is equal to

(A) $\frac{\pi}{2}$

(B) $\frac{\pi}{3}$

(D) $\frac{-3 \pi}{4}$

Solution: $\tan ^{-1}\left(\frac{x}{y}\right)-\tan ^{-1} \frac{x-y}{x+y} =\tan ^{-1}\left[\frac{\frac{x}{y}-\frac{x-y}{x+y}}{1+\left(\frac{x}{y}\right)\left(\frac{x-y}{x+y}\right)}\right]$