ExamBro
ExamBro
WBJEE · Maths · Differentiation

For \(-\frac{\pi}{2} < x < \frac{3 \pi}{2}, \quad\) the \(\quad\) value \(\quad\) of \(\frac{d}{d x}\left\{\tan ^{-1} \frac{\cos x}{1+\sin x}\right\}\) is equal to

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

Answer & Solution

Correct Answer

(B) \(-\frac{1}{2}\)

Step-by-step Solution

Detailed explanation

\(\frac{d}{d x}\left\{\tan ^{-1} \frac{\cos x}{1+\sin x}\right\}\) \(=\frac{1}{1+\left(\frac{\cos x}{1+\sin x}\right)^{2}} \frac{d}{d x} \frac{\cos x}{1+\sin x}\) \(=\frac{(1+\sin x)^{2}}{1+\sin ^{2} x+2 \sin x+\cos ^{2} x}\)…