Question:
Find the value of $\tan ^{-1} \sqrt{3}-\sec ^{-1}(-2)$ is equal to
(A) $\pi$
(B) $-\frac{\pi}{3}$
(C) $\frac{\pi}{3}$
(D) $\frac{2 \pi}{3}$
Solution:
Let $\tan ^{-1} \sqrt{3}=x$. Then, $\tan x=\sqrt{3}=\tan \frac{\pi}{3}$.
We know that the range of the principal value branch of $\tan ^{-1}$ is $\left(\frac{-\pi}{2}, \frac{\pi}{2}\right)$.
$\therefore \tan ^{-1} \sqrt{3}=\frac{\pi}{3}$
Let $\sec ^{-1}(-2)=y .$ Then, $\sec y=-2=-\sec \left(\frac{\pi}{3}\right)=\sec \left(\pi-\frac{\pi}{3}\right)=\sec \frac{2 \pi}{3}$.
We know that the range of the principal value branch of $\sec ^{-1}$ is $[0, \pi]-\left\{\frac{\pi}{2}\right\}$.
$\therefore \sec ^{-1}(-2)=\frac{2 \pi}{3}$
Hence, $\tan ^{-1}(\sqrt{3})-\sec ^{-1}(-2)=\frac{\pi}{3}-\frac{2 \pi}{3}=-\frac{\pi}{3}$