If \(y=\log \sqrt{\frac{1+\sin x}{1-\sin x}}\), then \(\frac{d y}{d x}\) at \(x=\frac{\pi}{3}\) is

\(y=\log \sqrt{\frac{1+\sin x}{1-\sin x}}=\frac{1}{2}\{\log (1+\sin x)-\log (1-\sin x)\} \)
\( \left\{\frac{1}{1+\sin x} \cdot \cos x-\frac{1}{1-\sin x} \cdot(-\cos x)\right\} \)
\( \Rightarrow \frac{d y}{d x}_{\text {at } x=\frac{\pi}{3}}=\frac{1}{2}\{\frac{1}{1+\sin \frac{\pi}{3}} \cos \frac{\pi}{3}-\frac{1}{1-\sin \frac{\pi}{3}} \cdot\) \(\left(-\cos \frac{\pi}{3}\right)\} \)
\( =\frac{1}{2} \frac{1}{2}\left\{\frac{1}{1+\frac{\sqrt{3}}{2}} \cdot \frac{1}{2}+\frac{1}{1-\frac{\sqrt{3}}{2}} \cdot \frac{1}{2}\right\} \)
\( =\frac{1}{2}\left\{\frac{1}{2+\sqrt{3}}+\frac{1}{2-\sqrt{3}}\right\} \)
\( =\frac{1}{2} \times \frac{4}{4-3}=2\)