WBJEE · Maths · Differentiation
If \(y=\tan ^{-1} \sqrt{\frac{1-\sin x}{1+\sin x}}\), then the value of \(\frac{d y}{d x}\) at \(x=\frac{\pi}{6}\) is
- A \(-\frac{1}{2}\)
- B \(\frac{1}{2}\)
- C 1
- D -1
Answer & Solution
Correct Answer
(A) \(-\frac{1}{2}\)
Step-by-step Solution
Detailed explanation
\text { Hints : } \begin{aligned} & y=\tan ^{-1} \sqrt{\frac{1-\cos \left(\frac{\pi}{2}-x\right)}{1+\cos \left(\frac{\pi}{2}-x\right)}} \\ & =\tan ^{-1} \sqrt{\frac{2 \sin ^2\left(\frac{\pi}{4}-\frac{x}{2}\right)}{2 \cos ^2\left(\frac{\pi}{4}-\frac{x}{2}\right)}}=\tan…
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