is equal to (A) \pi (B) 
(C) 
(D) 
The correct option is Option (B). $\tan ^{-1}(\sqrt{3})-\sec ^{-1}(-2)=\tan ^{-1}(\tan \pi / 3)-\sec ^{-1}(-\sec \pi / 3)$ $=\pi / 3-\sec ^{-1}(\sec (\pi-\pi / 3))$ $=\pi / 3-2 \pi / 3=-\pi /...
, then (A) 
(B) 
(C) 
(D) 
The correct option is Option (B) . Since, $\sin ^{-1} x=y$, The range of $\sin ^{-1}$ is $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ Therefore, $-\frac{\pi}{2} \leq y \leq \frac{\pi}{2}$
Assume $\cos ^{-1}\left(\frac{1}{2}\right)=x$. Then, $\cos x=\frac{1}{2}=\cos \left(\frac{\pi}{3}\right)$ $\cos ^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{3}$ Assume, $\sin...
${\cos e c^{-1}(-\sqrt{2})}$ assume, $y=\operatorname{cosec}^{-1}(-\sqrt{2})$ $\operatorname{cosec} y=-\sqrt{2}$ $\cos e c y=\cos e c \frac{-\pi}{4}$ Since$\operatorname{cosec}^{-1}$ is...
${\cot ^{-1}(\sqrt{3})}$ $y=\cot ^{-1}(\sqrt{3})$ $\cot y=\sqrt{3}$ $\cot y=\pi / 6$ Since $\cot ^{1}$ is $[0, \pi]$ Therefore, ${\cot ^{-1}(\sqrt{3})}$ is $\pi / 6$.
\sec ^{-1}\left(\frac{2}{\sqrt{3}}\right)$ $y=\sec ^{-1}\left(\frac{2}{\sqrt{3}}\right)$ $\sec y=2 / \sqrt{3}$ $\sec y=\sec \frac{\pi}{6}$ Since $\sec ^{-1}$ is $[0, \pi]$ Therefore, $\sec...
$\tan ^{-1}(-1)$ Let $y=\tan ^{-1}(-1)$ $\tan (y)=-1$ $\tan y=-\tan \pi / 4$ $\tan y=\tan \left(-\frac{\pi}{4}\right)$ Since tan $^{-1}$ is $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ Therefore,...
$y=\cos ^{-1}\left(\frac{-1}{2}\right)$ $\cos y=-1 / 2$ Since principle is $\cos ^{-1}$ is [0, \pi] Therefore, $\cos ^{-1}\left(\frac{-1}{2}\right)$ is $2 \pi / 3$.
$\tan ^{-1}(-\sqrt{3})$ Let $y=\tan ^{-1}(-\sqrt{3})$ $\tan y=-\tan \pi / 3$ or $\tan y=\tan (-\pi / 3)$ Since range is tan $^{-1}$ is $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ Therefore,$\tan...
Let $y=\operatorname{cosec}^{-1}(2)$ $\operatorname{Cosec} y=2$ since,$\operatorname{cosec} \pi / 6=2$ So $\operatorname{cosec} y=\operatorname{cosec} \pi / 6$ Since range is...
assuming, $y=\cos ^{-1}\left(\frac{\sqrt{3}}{2}\right)$ $\cos y=\cos \pi / 6 \quad(\operatorname{as} \cos \pi / 6=\sqrt{3} / 2)$ $y=\pi / 6$ Since range is $\cos ^{-1}$ is $[0, \pi]$ Therefore,...
Considering $y=\sin ^{-1}\left(-\frac{1}{2}\right)$ Solveing the above equation, we get $\sin \mathrm{y}=-1 / 2$ since, $\sin \pi / 6=1 / 2$ therefore, $\sin y=-\sin$ $\sin y=\sin...