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 \(y^{\prime}\left(\frac{\pi}{6}\right)=\)
- A \(-\frac{1}{4}\)
- B \(\frac{1}{6}\)
- C \(\frac{1}{4}\)
- D \(\frac{1}{2}\)
Answer & Solution
Correct Answer
(D) \(\frac{1}{2}\)
Step-by-step Solution
Detailed explanation
\(y=\tan ^{-1}\left(\sqrt{\frac{\left(\cos \frac{x}{2}+\sin \frac{x}{2}\right)^{2}}{\left(\cos \frac{x}{2}-\sin \frac{x}{2}\right)^{2}}}\right)\) \(y=\tan ^{-1}\left(\frac{\cos \frac{x}{2}+\sin \frac{x}{2}}{\cos \frac{x}{2}-\sin \frac{x}{2}}\right)\)
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