Let $y=tan^{-1}\begin{Bmatrix}\frac{\sqrt{1+sin\, x}+\sqrt{1-sin\, x}}{\sqrt{1+sin\, x}-\sqrt{1-sin\, x}}\end{Bmatrix}, 0 < x< \frac{\pi}{2}$ then $\frac{dy}{dx}=$
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → $-\frac{1}{2}$
$y=\tan^{-1}\left\{\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}×\frac{\sqrt{1+\sin x}-\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right\}$
$y=\tan^{-1}\left\{\frac{2\sin x}{1+\sin x+1-\sin x-2\sqrt{1-\sin^2x}}\right\}$
$y=\tan^{-1}\left\{\frac{2\sin x}{2(1-\cos x)}\right\}$
$y=\tan^{-1}\left(\frac{2\sin\frac{x}{2}\cos\frac{x}{2}}{2\sin^2\frac{x}{2}}\right)=\tan^{-1}(\cot\frac{x}{2})$
$y=\frac{π}{2}-\frac{x}{2}⇒\frac{dy}{dx}=-\frac{1}{2}$