Trigonometric Functions — Maths STD 12 Science — Question

$\therefore \tan \alpha=\sqrt{3}=\tan \frac{\pi}{3}$

$\therefore \alpha=\frac{\pi}{3} \quad \cdots\left[\because \frac{-\pi}{2}<\frac{\pi}{3}<\frac{\pi}{2}\right]$

$\therefore \tan ^{-1}(\sqrt{3})=\frac{\pi}{3}$$\ldots(1)$

Let $\sec ^{-1}(-2)=\beta$, where $0 \leqslant \beta \leqslant \pi, \beta \neq \frac{\pi}{2}$

$\therefore \sec \beta=-2=-\sec \frac{\pi}{3}$

$\therefore \sec \beta=\sec \left(\pi-\frac{\pi}{3}\right) \ldots[\because \sec (\pi-\theta)=-\sec \theta]$

$\therefore \sec \beta=\sec \frac{2 \pi}{3}$

$\therefore \beta=\frac{2 \pi}{3} \quad \ldots\left[\left[\because 0 \leqslant \frac{2 \pi}{3} \leqslant \pi\right]\right.$

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

$\ldots(2)$

$\begin{aligned} & \therefore \tan ^{-1} \sqrt{3}-\sec ^{-1}(-2) \\ & =\frac{\pi}{3}-\frac{2 \pi}{3} \ldots[B y(1) \text { and }(2)] \\ & =-\frac{\pi}{3} .\end{aligned}$