Find the principal values of each of the following:

Solution:

(i) Let $\cot ^{-1}(-\sqrt{3})=y$

Then,

$\cot y=-\sqrt{3}$

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

Thus,

$\cot y=-\sqrt{3}=\cot \left(\frac{5 \pi}{6}\right)$

$\Rightarrow y=\frac{5 \pi}{6} \in(0, \pi)$

Hence, the principal value of $\cot ^{-1}(-\sqrt{3})$ is $\frac{5 \pi}{6}$.

(ii) Let $\cot ^{-1}(\sqrt{3})=y$

Then,

$\cot y=\sqrt{3}$

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

Thus,

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

$\Rightarrow y=\frac{\pi}{6} \in(0, \pi)$

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

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

Then,

$\cot y=-\frac{1}{\sqrt{3}}$

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

Thus,

$\cot y=-\frac{1}{\sqrt{3}}=\cot \left(\frac{2 \pi}{3}\right)$

$\Rightarrow y=\frac{2 \pi}{3} \in(0, \pi)$

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

(iv)

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

Then,

$\cot y=\tan \frac{3 \pi}{4}$

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

Thus,

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

$\Rightarrow y=\frac{3 \pi}{4} \in(0, \pi)$

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