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Continuity and Differentiability — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsContinuity and Differentiability3 Marks
Question
Differentiate the function $\cot ^{-1}\left[\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right], 0<x<\frac{\pi}{2}$ w.r.t. x.
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Answer

$y = {\cot ^{ - 1}}\left[ {\frac{{\sqrt {{{\left( {\cos \frac{x}{2} + \sin \frac{x}{2}} \right)}^2}} + \sqrt {{{\left( {\cos \frac{x}{2} - \sin \frac{x}{2}} \right)}^2}} }}{{\sqrt {{{\left( {\cos \frac{x}{2} + \sin \frac{x}{2}} \right)}^2}} - \sqrt {{{\left( {\cos \frac{x}{2} - \sin \frac{x}{2}} \right)}^2}} }}} \right]$   $\left[ \because\sqrt {1 \pm \sin x} = \sqrt {{{\left( {\cos \frac{x}{2} \pm \sin \frac{x}{2}} \right)}^2}} \right] $

$= {\cot ^{ - 1}}\frac{{\cos \frac{x}{2} + \sin \frac{x}{2} + \cos \frac{x}{2} - \sin \frac{x}{2}}}{{\cos \frac{x}{2} + \sin \frac{x}{2} - \cos \frac{x}{2} + \sin \frac{x}{2}}}$

$ = {\cot ^{ - 1}}\left( {\frac{{2\cos \frac{x}{2}}}{{2\sin \frac{x}{2}}}} \right)$

$ = {\cot ^{ - 1}}\left( {\cot \frac{x}{2}} \right)$

$ = \frac{x}{2}$

$y = \frac{x}{2}$

$\frac{{dy}}{{dx}} = \frac{1}{2}$

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