# For the principal values, evaluate each of the following:

Question:

For the principal values, evaluate each of the following:

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

(ii) $\tan ^{-1}\left\{2 \sin \left(4 \cos ^{-1} \frac{\sqrt{3}}{2}\right)\right\}$

Solution:

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

$=\tan ^{-1}\left\{\tan \left(-\frac{\pi}{4}\right)\right\}+\cos ^{-1}\left(\cos \frac{3 \pi}{4}\right)$                    $\left[\because\right.$ Range of $\tan$ is $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right) ;-\frac{\pi}{4} \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ and range of cosine is $\left.[0, \pi] ; \frac{3 \pi}{4} \in[0, \pi]\right]$

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

$=\frac{\pi}{2}$

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

(ii)

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

$=\tan ^{-1}\left\{2 \sin \left[4 \times \frac{\pi}{6}\right]\right\}$

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

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

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

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

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