The derivative of y = tan^{-1} left[ frac{sqr | vedclass

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(D) Given $y = 	an^{-1} \left[ \frac{\sqrt{1 + \sin x} + \sqrt{1 - \sin x}}{\sqrt{1 + \sin x} - \sqrt{1 - \sin x}} ight]$. We know that $1 + \sin x

Vedclass · Accepted Answer

$\pm \frac{1}{2}$

Vedclass · Answer

(D) Given $y = 	an^{-1} \left[ \frac{\sqrt{1 + \sin x} + \sqrt{1 - \sin x}}{\sqrt{1 + \sin x} - \sqrt{1 - \sin x}} ight]$. We know that $1 + \sin x = (\cos \frac{x}{2} + \sin \frac{x}{2})^2$ and $1 - \sin x = (\cos \frac{x}{2} - \sin \frac{x}{2})^2$. Thus,$\sqrt{1 + \sin x} = |\cos \frac{x}{2} + \sin \frac{x}{2}|$ and $\sqrt{1 - \sin x} = |\cos \frac{x}{2} - \sin \frac{x}{2}|$. Substituting these into the expression for $y$: $y = 	an^{-1} \left[ \frac{(\cos \frac{x}{2} + \sin \frac{x}{2}) + (\cos \frac{x}{2} - \sin \frac{x}{2})}{(\cos \frac{x}{2} + \sin \frac{x}{2}) - (\cos \frac{x}{2} - \sin \frac{x}{2})} ight]$ (assuming $0 $y = 	an^{-1} \left( \frac{2 \cos \frac{x}{2}}{2 \sin \frac{x}{2}} ight) = 	an^{-1} (\cot \frac{x}{2}) = 	an^{-1} \left( 	an (\frac{\pi}{2} - \frac{x}{2}) ight) = \frac{\pi}{2} - \frac{x}{2}$. Differentiating with respect to $x$: $\frac{dy}{dx} = \frac{d}{dx} (\frac{\pi}{2} - \frac{x}{2}) = -\frac{1}{2}$. Depending on the interval of $x$,the derivative can be $\pm \frac{1}{2}$. Therefore,the correct option is $(D)$.

The derivative of $y = \tan^{-1} \left[ \frac{\sqrt{1 + \sin x} + \sqrt{1 - \sin x}}{\sqrt{1 + \sin x} - \sqrt{1 - \sin x}} \right]$ with respect to $x$ is equal to

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