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  4. If 0< x< (π/2) then cot -1 (√1+ sin x+√1- sin x/√1+ sin x-√1- sin x) =

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Q. If $0< x< \frac{\pi}{2}$ then $\cot ^{-1}\left\{\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right\}=$

KCETKCET 2022

Solution:

$\sqrt{1+\sin x}=\sqrt{\cos ^{2} \frac{x}{2}+\sin ^{2} \frac{x}{2}+2 \cdot \sin \frac{x}{2} \cdot \cos \frac{x}{2}}$
$=\sqrt{\left(\cos \frac{x}{2}+\sin \frac{x}{2}\right)^{2}}=\cos \frac{x}{2}+\sin \frac{x}{2}$
Similarly $\sqrt{1-\sin x}=\cos \frac{x}{2}-\sin \frac{x}{2}$
$\left(\right.$ Since $0< x< \frac{\pi}{2}$ we have, $\left.\cos \frac{x}{2}>\sin \frac{x}{2}\right)$
$\therefore \cot ^{-1}\left\{\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right\}$
$=\cot ^{-1} \frac{\cos \left(\frac{x}{2}\right)+\sin \left(\frac{x}{2}\right)+\cos \left(\frac{x}{2}\right)-\sin \left(\frac{x}{2}\right)}{\cos \left(\frac{x}{2}\right)+\sin \left(\frac{x}{2}\right)-\left(\cos \left(\frac{x}{2}\right)-\sin \left(\frac{x}{2}\right)\right)}$
$=\cot ^{-1} \cot \frac{x}{2}=\frac{x}{2}$