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MHT CET · Maths · Inverse Trigonometric Functions

If \(y=\tan ^{-1}\left(\sqrt{\frac{1+\sin x}{1-\sin x}}\right), \quad 0 \leqslant x < \frac{\pi}{2}\), then \(y^{\prime}\left(\frac{\pi}{6}\right)=\)

  1. A \(-\frac{1}{4}\)
  2. B \(\frac{1}{6}\)
  3. C \(\frac{1}{4}\)
  4. D \(\frac{1}{2}\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(\frac{1}{2}\)

Step-by-step Solution

Detailed explanation

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