Question

Differentiate

(i) ${\tan }^{-1}\left\{\frac{x}{1+\sqrt{1-x^2}}\right\},-1<x<1$

(ii) ${\tan }^{-1}\left(\frac{1+\cos x}{\sin x}\right)$

Answer 1

(i)
$$\begin{gathered} \mathrm{Let},y={\tan }^{-1}\left\{\frac{x}{1+\sqrt{1-x^2}}\right\} \\ \mathrm{Put}x=\sin \theta \\ y={\tan }^{-1}\left\{\frac{\sin \theta }{1+\sqrt{1-{\sin }^2\theta }}\right\} \\ y={\tan }^{-1}\left(\frac{\sin \theta }{1+\cos \theta }\right) \\ y={\tan }^{-1}\left\{\frac{2\sin {\displaystyle \frac{\theta }{2}}\cos {\displaystyle \frac{\theta }{2}}}{2{\cos }^2{\displaystyle \frac{\theta }{2}}}\right\} \\ y={\tan }^{-1}\left\{\tan \frac{\theta }{2}\right\}...\left(\mathrm{i}\right) \\ \mathrm{Here},-1<x<1 \\ \Rightarrow -1<\sin \theta <1 \\ \Rightarrow -\frac{\mathrm{\pi }}{2}<\theta <\frac{\mathrm{\pi }}{2} \\ \Rightarrow -\frac{\mathrm{\pi }}{4}<\frac{\theta }{2}<\frac{\mathrm{\pi }}{4} \\ \mathrm{So},\mathrm{from}\mathrm{equation}\left(\mathrm{i}\right), \\ y=\frac{\theta }{2}\left[\mathrm{Since},{\tan }^{-1}\left(\tan \theta \right)=\theta ,if\theta \in \left[-\frac{\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{2}\right]\right] \\ y=\frac{1}{2}{\sin }^{-1}x\left[\mathrm{Since},x=\sin \theta \right] \\ \mathrm{Differentiating}\mathrm{it}\mathrm{with}\mathrm{respect}\mathrm{to}x, \\ \frac{dy}{dx}=\frac{1}{2\sqrt{1-x^2}} \end{gathered}$$

(ii)
$$\begin{gathered} \mathrm{Let}y={\tan }^{-1}\left(\frac{1+\cos x}{\sin x}\right).\mathrm{Then}, \\ \Rightarrow y={\tan }^{-1}\left(\frac{2{\cos }^2{\displaystyle \frac{x}{2}}}{2\sin {\displaystyle \frac{x}{2}}.\cos {\displaystyle \frac{x}{2}}}\right) \\ \Rightarrow y={\tan }^{-1}\left(\cot \frac{x}{2}\right) \\ \Rightarrow y={\tan }^{-1}\left\{\tan \left(\frac{\mathrm{\pi }}{2}-\frac{x}{2}\right)\right\} \\ \Rightarrow y=\frac{\mathrm{\pi }}{2}-\frac{x}{2} \\ \therefore \frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}=0-\frac{1}{2}=-\frac{1}{2} \end{gathered}$$