SOLUTION : (i) Let $\sin ^{-1} x=\theta$. Then $x=\sin \theta$ and $-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}$
$\therefore \quad \cos \theta=\sqrt{1-\sin ^2 \theta}=\sqrt{1-x^2} \\$
$\therefore \quad \cos ^{-1} \sqrt{1-x^2}=\cos ^{-1}(\cos \theta)=$
$-\theta \text { if }-\frac{\pi}{2} \leq \theta<0 \\$
$\theta \text { if } 0 \leq \theta \leq \frac{\pi}{2}$
$-\sin ^{-1} x \text { if }-1 \leq x<0 \\$
$\sin ^{-1} x \text { if } 0 \leq x \leq 1$
$\therefore\quad \sin ^{-1} x$=
$-\cos ^{-1} \sqrt{1-x^2} \text { if }-1 \leq x<0 \\$
$\cos ^{-1} \sqrt{1-x^2} \text { if } 0 \leq x \leq 1$
(ii) Let $\sin ^{-1} x=\theta$. Then $x=\sin \theta$ and $-\frac{\pi}{2}<\theta<\frac{\pi}{2} \quad$ Note: $\theta \neq-\frac{\pi}{2}$, $\frac{\pi}{2}$ because $\left.x \neq \pm 1\right\}$
$\therefore \quad \tan \theta=\frac{\sin \theta}{\cos \theta}=\frac{\mathrm{x}}{\sqrt{1-\mathrm{x}^2}} \quad \therefore \quad \tan ^{-1} \frac{\mathrm{x}}{\sqrt{1-\mathrm{x}^2}}=\tan ^{-1}(\tan \theta)=\theta=\sin ^{-1} \mathrm{x}$
Thus $\sin ^{-1} x=\tan ^{-1} \frac{x}{\sqrt{1-x^2}}$, for all $x \in(-1,1)$
(iii) Let $\sin ^{-1} x=\theta$. Then $x=\sin \theta$ and $-\frac{\pi}{2} \leq \theta<0$ or $0<\theta \leq \frac{\pi}{2}$ \{Note: $\theta \neq 0$, because $\left.x \neq 0\right\}$
$\therefore \quad \cot \theta=\frac{\sqrt{1-x^2}}{x} \\$
$\therefore \quad \cot ^{-1} \frac{\sqrt{1-x^2}}{x}=\cot ^{-1}(\cot \theta)=$
$\theta+\pi \text { if }-\frac{\pi}{2} \leq \theta<0 \\$
$\theta \text { if } 0<\theta \leq \frac{\pi}{2}$
$\pi+\sin ^{-1} x \text { if }-1 \leq x<0 \\$
$\sin ^{-1} x \text { if } 0<x \leq 1$
Thus $\sin ^{-1} x=\left\{\begin{array}{ll}\cot ^{-1} \frac{\sqrt{1-x^2}}{x}-\pi & \text { if }-1 \leq x<0 \\ \cot ^{-1} \frac{\sqrt{1-x^2}}{x} & \text { if } 0<x \leq 1\end{array}\right.$
