# Find the principal values of each of the following:

Question:

Find the principal values of each of the following:

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

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

(iii) $\tan ^{-1}\left(\cos \frac{\pi}{2}\right)$

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

Solution:

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

Then,

$\tan y=\frac{1}{\sqrt{3}}$

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

Thus,

$\tan y=\frac{1}{\sqrt{3}}=\tan \left(\frac{\pi}{6}\right)$

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

Hence, the principal value of $\tan ^{-1}\left(\frac{1}{\sqrt{3}}\right)$ is $\frac{\pi}{6}$.\

(ii) We have $\tan ^{-1}\left(-\frac{1}{\sqrt{3}}\right)=-\tan ^{-1}\left(\frac{1}{\sqrt{3}}\right)$                                 $\left[\because \tan ^{-1}(-x)=-\tan ^{-1} x\right]$

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

Then,

$\tan y=\frac{1}{\sqrt{3}}$

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

Thus,

$\tan y=\frac{1}{\sqrt{3}}=\tan \left(\frac{\pi}{6}\right)$

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

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

$=-y$

$=-\frac{\pi}{6} \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$

Hence, the principal value of $\tan ^{-1}\left(-\frac{1}{\sqrt{3}}\right)$ is $-\frac{\pi}{6}$.

(iii) Let $\tan ^{-1}\left(\cos \frac{\pi}{2}\right)=y$

Then,

$\tan y=\cos \frac{\pi}{2}$

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

Thus,

$\tan y=\cos \frac{\pi}{2}=0=\tan (0)$

$\Rightarrow y=0 \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$

Hence, the principal value of $\tan ^{-1}\left(\cos \frac{\pi}{2}\right)$ is 0 .

(iv) Let $\tan ^{-1}\left(2 \cos \frac{2 \pi}{3}\right)=y$

Then,

$\tan y=2 \cos \frac{2 \pi}{3}$

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

Thus,

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

$\Rightarrow y=-\frac{\pi}{4} \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$

Hence, the principal value of $\tan ^{-1}\left(2 \cos \frac{2 \pi}{3}\right)$ is $-\frac{\pi}{4}$.