Let \(y=\tan^{-1}\left\{\displaystyle\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right\}\), \(0\lt x\lt\displaystyle\frac{\pi}{2}\) then \(\displaystyle\frac{dy}{dx}=\)
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(-\displaystyle\frac{1}{2}\)
The bracket simplifies before differentiating. Using \(\sin x=2\sin\frac{x}{2}\cos\frac{x}{2}\) and \(1=\cos^2\frac{x}{2}+\sin^2\frac{x}{2}\):
\( \displaystyle 1+\sin x=\left(\cos\frac{x}{2}+\sin\frac{x}{2}\right)^2,\quad 1-\sin x=\left(\cos\frac{x}{2}-\sin\frac{x}{2}\right)^2 \)
For \(0\lt x\lt\frac{\pi}{2}\), \(0\lt\frac{x}{2}\lt\frac{\pi}{4}\), so \(\cos\frac{x}{2}\gt\sin\frac{x}{2}\gt 0\) and both brackets are positive:
\( \displaystyle \sqrt{1+\sin x}=\cos\frac{x}{2}+\sin\frac{x}{2},\quad \sqrt{1-\sin x}=\cos\frac{x}{2}-\sin\frac{x}{2} \)
Then
\( \displaystyle \frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}=\frac{2\cos\frac{x}{2}}{2\sin\frac{x}{2}}=\cot\frac{x}{2} \)
\( \displaystyle =\tan\left(\frac{\pi}{2}-\frac{x}{2}\right) \)
The angle \(\frac{\pi}{2}-\frac{x}{2}\) lies in \(\left(\frac{\pi}{4}, \frac{\pi}{2}\right)\), inside the principal value range of \(\tan^{-1}\), so
\( \displaystyle y=\frac{\pi}{2}-\frac{x}{2} \)
\( \displaystyle \frac{dy}{dx}=-\frac{1}{2} \)