2. inverse trigonometric function — ગણિત ધોરણ 12 સાયન્સ — Question

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 $[0, \pi]-\left\{\frac{\pi}{2}\right\}$

$\therefore \sec ^{-1}(-2)=\frac{2 \pi}{3}$

Thus, $\tan ^{-1}(\sqrt{3})-\sec ^{-1}(-2)$

$=\frac{\pi}{3}-\frac{2 \pi}{3}=-\frac{\pi}{3}$