Evaluate each of the following:

Solution:

(i) Let $\sin ^{-1}\left(-\frac{1}{2}\right)=y$

Then,

$\sin y=-\frac{1}{2}$

We know that the range of the principal value branch is $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$.

Thus,

$\sin y=-\frac{1}{2}=\sin \left(-\frac{\pi}{6}\right)$

$\Rightarrow y=-\frac{\pi}{6} \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$

Now,

Let $\cos ^{-1}\left(-\frac{1}{2}\right)=z$

Then,

$\cos z=-\frac{1}{2}$

We know that the range of the principal value branch is $[0, \pi]$.

Thus,

$\cos z=-\frac{1}{2}=\cos \left(\frac{2 \pi}{3}\right)$

$\cos z=-\frac{1}{2}=\cos \left(\frac{2 \pi}{3}\right)$

$\Rightarrow z=\frac{2 \pi}{3} \in[0, \pi]$

So,

$\tan ^{-1} 1+\cos ^{-1}\left(-\frac{1}{2}\right)+\sin ^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{4}+\frac{2 \pi}{3}-\frac{\pi}{6}=\frac{3 \pi}{4}$

$\therefore \tan ^{-1} 1+\cos ^{-1}\left(-\frac{1}{2}\right)+\sin ^{-1}\left(\frac{1}{2}\right)=\frac{3 \pi}{4}$

(ii)

$\tan ^{-1}\left(-\frac{1}{\sqrt{3}}\right)+\tan ^{-1}(-\sqrt{3})+\tan ^{-1}\left(\sin \left(-\frac{\pi}{2}\right)\right)$

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

$=\tan ^{-1}\left(-\frac{1}{\sqrt{3}}\right)+\tan ^{-1}(-\sqrt{3})+\tan ^{-1}(-1)$

$=-\tan ^{-1}\left(\frac{1}{\sqrt{3}}\right)-\tan ^{-1}(\sqrt{3})-\tan ^{-1}(1)$

$=-\tan ^{-1}\left(\tan \frac{\pi}{6}\right)-\tan ^{-1}\left(\frac{\pi}{3}\right)-\tan ^{-1}\left(\frac{\pi}{4}\right)$

$=-\frac{\pi}{6}-\frac{\pi}{3}-\frac{\pi}{4}$

$=-\frac{3 \pi}{4}$

(iii)

$\tan ^{-1}\left(\tan \frac{5 \pi}{6}\right)+\cos ^{-1}\left\{\cos \left(\frac{13 \pi}{6}\right)\right\}$

$=\tan ^{-1}\left\{\tan \left(\pi-\frac{5 \pi}{6}\right)\right\}+\cos ^{-1}\left\{\cos \left(2 \pi+\frac{\pi}{6}\right)\right\}$

$=\tan ^{-1}\left\{-\tan \left(\frac{\pi}{6}\right)\right\}+\cos ^{-1}\left\{\cos \left(\frac{\pi}{6}\right)\right\}$

$=-\tan ^{-1}\left\{\tan \left(\frac{\pi}{6}\right)\right\}+\cos ^{-1}\left\{\cos \left(\frac{\pi}{6}\right)\right\}$

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

$=0$