(a)OR(b)

Simplify $\displaystyle \sin ^{-1}\left(\frac{\mathrm{x}}{\sqrt{1+\mathrm{x}^{2}}}\right)$.

Put $\displaystyle \mathrm{x}=\tan \theta \Rightarrow \theta=\tan ^{-1} \mathrm{x}$ Now $\displaystyle \sin ^{-1}\left(\frac{\mathrm{x}}{\sqrt{1+\mathrm{x}^{2}}}\right)$ $\displaystyle =\sin ^{-1}\left(\frac{\tan \theta}{\sec \theta}\right)=\sin ^{-1}(\sin \theta)$ $\displaystyle =\theta=\tan ^{-1} \mathrm{x}$ Here $\displaystyle -1 \leq \sqrt{\mathrm{x}-1} \leq 1$ $\displaystyle \Rightarrow 0 \leq \mathrm{x}-1 \leq 1 \Rightarrow 1 \leq \mathrm{x} \leq 2$ Hence, domain is $\displaystyle \mathrm{x} \in[1,2]$