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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

  1. A \(-\frac{1}{2}\)
  2. B \(\frac{1}{2}\)
  3. C 1
  4. D -1
Verified Solution

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…