(A) ${ }^{1 / 2}$ (B) $1 / 3$ (C) ${ }^{1 / 4}$ (D) 1 The corect option is OPTION(D) Reason: solving for: $\sin ^{-1}\left(-\frac{1}{2}\right)$ we get $=\sin (\pi / 2)$ which is equal to...

(A) $7 \mathrm{\pi} / 6$ (B) $5 \mathrm{~m} / 6$ (C) $\pi / 3$ (D) $\pi / 6$ The correct option is Option (B)because, $\cos ^{-1}\left(\cos \frac{7 \pi}{6}\right)=\cos ^{-1}\left(\cos \left(2...

$\tan \left(\sin ^{-1}\left(\frac{3}{5}\right)+\cot ^{-1} \frac{3}{2}\right)$ replacing $\sin ^{-1}\left(\frac{3}{5}\right)by x$ and $\cot ^{-1}\left(\frac{3}{2}\right)by y$ Or $\sin (x)=3 / 5$ and...

$\tan ^{-1}\left(\tan \left(\frac{3 \pi}{4}\right)\right)$splitting $\frac{3 \pi}{4}$ as $\frac{(4 \pi-\pi)}{4}$ or $\pi-\frac{\pi}{4}$ and substituting the same given equation we get, $\tan...

$\sin ^{-1}\left(\sin \left(\frac{2 \pi}{3}\right)\right)$ splitting $\frac{2 \pi}{3}$ as $\frac{(3 \pi-\pi)}{3}$ or $\pi-\frac{\pi}{3}$ and substitute the same in given equation we get,$\sin...

usig the identity:: $\tan ^{-1} x+\tan ^{-1} y=\tan ^{-1} \frac{x+y}{1-x y}$we conclude: $\tan ^{-1}...

since, $\sin 90$ $=\sin \pi / 2=1$ using the above we get,, $\sin ^{-1} \frac{1}{5}+\cos ^{-1} x=\frac{\pi}{2}$ $\cos ^{-1} x=\frac{\pi}{2}-\sin ^{-1} \frac{1}{5}$ Using identity: $\sin ^{-1} t+\cos...

replacing x by $\tan \theta$ and y by $\tan \Phi$, we get, $\tan \frac{1}{2}\left[\sin ^{-1} \frac{2 \tan \theta}{1+\tan ^{2} \theta}+\cos ^{-1} \frac{1-\tan ^{2} \phi}{1+\tan ^{2} \phi}\right]$...

$\cot \left(\tan ^{-1} a+\cot ^{-1} a\right)=\cot \pi / 2=0$ formula used is: $\tan ^{-1} a+\cot ^{-1} a=\pi / 2$

$=\tan ^{-1}\left[2 \cos \left(2 \sin ^{-1} \sin \frac{\pi}{6}\right)\right]$ 8$=\tan ^{-1}\left[2 \cos \left(2 \times \frac{\pi}{6}\right)\right]$ $=\tan ^{-1}(2 \cos \pi / 3)$$=\tan ^{-1}(2 \times...

dividing both numerator and denominator by a $^{\wedge} 3$ we get,$\tan ^{-1}\left(\frac{3\left(\frac{x}{a}\right)-\left(\frac{x}{a}\right)^{3}}{1-3\left(\frac{x}{a}\right)^{2}}\right)$ Putting $x /...

replacing x by $a \sin \theta$, which is equal to $\sin \theta=x / a$ and $\theta=\sin ^{-1}(x / a)$ Substituting the values in the given function, we get $\tan ^{-1}...

dividing both numerator and denominator by $\cos x$,we get $\tan ^{-1}\left(\frac{\frac{\operatorname{cox}(x)}{\cos (x)}-\frac{\sin (x)}{\cos (x)}}{\frac{\cos (x)}{\cos (x)}+\frac{\sin (x)}{\cos...

putting, $x=\sec \theta$, which means $\theta=\sec ^{-1} x$ $\tan ^{-1} \frac{1}{\sqrt{x^{2}-1}}=\tan ^{-1} \frac{1}{\sqrt{\sec ^{2} \theta-1}}$ $=\tan ^{-1} \frac{1}{\sqrt{\tan ^{2} \theta}}$...

putting $x=\tan \theta$ which means $\theta=\tan ^{-1} x$ $\tan ^{-1} \frac{\sqrt{1+x^{2}}-1}{x}=\tan ^{-1}\left(\frac{\sqrt{1+\tan ^{2} \theta}-1}{\tan \theta}\right)$ $=\tan ^{-1}\left(\frac{\sec...

identity used is: $2 \tan ^{-1} x=\tan ^{-1} \frac{2 x}{1-x^{2}}$ LHS $=2 \tan ^{-1} \frac{1}{2}+\tan ^{-1} \frac{1}{7}$ $=\tan ^{-1} \frac{2 \times \frac{1}{2}}{1-\left(\frac{1}{2}\right)^{2}}+\tan...

$3 \cos ^{-1} x=\cos ^{-1}\left(4 x^{3}-3 x\right), x \in\left[\frac{1}{2}, 1\right]$ identity used is: $\cos 3 \theta=4 \cos ^{3} \theta-3 \cos \theta$ $\operatorname{Putting} x=\cos \theta$...