Trigonometric Functions — Maths STD 12 Science — Question

$\therefore \operatorname{cosec} \alpha=-\sqrt{2}=-\operatorname{cosec} \frac{\pi}{4}$

$\therefore \operatorname{cosec} \alpha=\operatorname{cosec}\left(-\frac{\pi}{4}\right)$$\ldots[\because \operatorname{cosec}(-\theta)=-\operatorname{cosec} \theta]$

$\therefore \alpha=-\frac{\pi}{4} \quad \quad \ldots\left[\frac{-\pi}{2} \leqslant \frac{-\pi}{4} \leqslant \frac{\pi}{2}\right]$

$\therefore \operatorname{cosec}^{-1}(-\sqrt{2})=-\frac{\pi}{4}$$\ldots(1)$

Let $\cot ^{-1}(\sqrt{3})=\beta$, where $0<\beta<\pi$

$\therefore \cot \beta=\sqrt{3}=\cot \frac{\pi}{6}$

$\therefore \beta=\frac{\pi}{6} \quad \cdots\left[\because 0<\frac{\pi}{6}<\pi\right]$

$\therefore \cot ^{-1}(\sqrt{3})=\frac{\pi}{6}$

$\ldots(2)$

$\therefore \operatorname{cosec}^{-1}(-\sqrt{2})+\cot ^{-1}(\sqrt{3})$

$=-\frac{\pi}{4}+\frac{\pi}{6} \quad \ldots .[$ By (1) and (2)]

$=\frac{-3 \pi+2 \pi}{12}=-\frac{\pi}{12}$