Question

If $f\left(x\right)={\tan }^{-1}\sqrt{\frac{1+\sin x}{1-\sin x}},0\le x\le \pi /2,\mathrm{then}f\text{'}\left(\pi /6\right)\mathrm{is}$
(a) − 1/4
(b) − 1/2
(c) 1/4
(d) 1/2

Answer 1

(d) 1/2

$$\begin{gathered} \mathrm{Let}y={\tan }^{-1}\left\{\sqrt{\frac{1+\sin x}{1-\sin x}}\right\} \\ \Rightarrow y={\tan }^{-1}\left\{\sqrt{\frac{1-\cos \left({\displaystyle \frac{\mathrm{\pi }}{2}}+x\right)}{1+\cos \left({\displaystyle \frac{\mathrm{\pi }}{2}}+x\right)}}\right\} \\ \Rightarrow y={\tan }^{-1}\left\{\sqrt{\frac{2{\sin }^2\left({\displaystyle \frac{\mathrm{\pi }}{4}}+{\displaystyle \frac{x}{2}}\right)}{2{\cos }^2\left(\frac{\mathrm{\pi }}{4}+\frac{x}{2}\right)}}\right\} \\ \Rightarrow y={\tan }^{-1}\left\{\tan \left(\frac{\mathrm{\pi }}{4}+\frac{x}{2}\right)\right\}=\frac{\mathrm{\pi }}{4}+\frac{x}{2} \\ \therefore \frac{dy}{dx}=\frac{1}{2} \end{gathered}$$