MHT CET · Maths · Differentiation
\(\frac{\mathrm{d}}{\mathrm{d} x}\left(\log \sqrt{\frac{1+\sin x}{1-\sin x}}\right)=\)
- A \(\cos ^2 x\)
- B \(\sec ^2 x\)
- C \(\cos x\)
- D \(\sec x\)
Answer & Solution
Correct Answer
(D) \(\sec x\)
Step-by-step Solution
Detailed explanation
\(\frac{d\left\{\log \sqrt{\frac{1+\sin x}{1-\sin x}}\right\}}{\mathrm{d} x}=\frac{\mathrm{d}\left\{\frac{1}{2} \log \left(\frac{1+\sin x}{1-\sin x}\right)\right\}}{\mathrm{d} x} \)
\( =\frac{1}{2} \times \frac{1}{\frac{1+\sin x}{1-\sin x}} \times \frac{(1-\sin x)(0+\cos x)-(1+\sin x)(0-\cos x)}{(1-\sin x)^2} \)
\( =\frac{1}{2} \times \frac{2 \cos x}{(1+\sin x)(1-\sin x)} \)
\( =\frac{\cos x}{1-\sin ^2 x}=\frac{\cos x}{\cos ^2 x}=\sec x\)
\( =\frac{1}{2} \times \frac{1}{\frac{1+\sin x}{1-\sin x}} \times \frac{(1-\sin x)(0+\cos x)-(1+\sin x)(0-\cos x)}{(1-\sin x)^2} \)
\( =\frac{1}{2} \times \frac{2 \cos x}{(1+\sin x)(1-\sin x)} \)
\( =\frac{\cos x}{1-\sin ^2 x}=\frac{\cos x}{\cos ^2 x}=\sec x\)
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