$\frac{\mathrm{d}}{\mathrm{d} x}\left(\log \sqrt{\frac{1+\sin x}{1-\sin x}}\right)=$
$\frac{\mathrm{d}}{\mathrm{d} x}\left(\log \sqrt{\frac{1+\sin x}{1-\sin x}}\right)=$
- $\cos ^2 x$
- $\sec ^2 x$
- $\cos x$
- $\sec x$
Solution
$\begin{aligned} & \frac{d\left\{\log \sqrt{\frac{1+\sin x}{1-\sin x}}\right\}}{\mathrm{d} x}=\frac{\mathrm{d}\left\{\frac{1}{2} \log \left(\frac{1+\sin x}{1-\sin x}\right)\right\}}{\mathrm{d} x} \\ & =\frac{1}{2} \times \frac{1}{\frac{1+\sin x}{1-\sin x}} \times \frac{(1-\sin x)(0+\cos x)-(1+\sin x)(0-\cos x)}{(1-\sin x)^2} \\ & =\frac{1}{2} \times \frac{2 \cos x}{(1+\sin x)(1-\sin x)} \\ & =\frac{\cos x}{1-\sin ^2 x}=\frac{\cos x}{\cos ^2 x}=\sec x\end{aligned}$
Asked in: MHT CET 2022 (08 Aug Shift 2)
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