If y = tan^{-1} sqrt{frac{a - x}{a + x}},then | vedclass

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(D) Given $y = 	an^{-1} \sqrt{\frac{a - x}{a + x}}$.Let $x = a \cos 	heta$,then $	heta = \cos^{-1}(\frac{x}{a})$.Substituting $x$ in the expression

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(D) Given $y = 	an^{-1} \sqrt{\frac{a - x}{a + x}}$.Let $x = a \cos 	heta$,then $	heta = \cos^{-1}(\frac{x}{a})$.Substituting $x$ in the expression:$y = 	an^{-1} \sqrt{\frac{a - a \cos 	heta}{a + a \cos 	heta}} = 	an^{-1} \sqrt{\frac{1 - \cos 	heta}{1 + \cos 	heta}}$Using trigonometric identities $1 - \cos 	heta = 2 \sin^2(\frac{	heta}{2})$ and $1 + \cos 	heta = 2 \cos^2(\frac{	heta}{2})$:$y = 	an^{-1} \sqrt{\frac{2 \sin^2(\frac{	heta}{2})}{2 \cos^2(\frac{	heta}{2})}} = 	an^{-1} \sqrt{	an^2(\frac{	heta}{2})} = \frac{	heta}{2}$Substituting $	heta$ back:$y = \frac{1}{2} \cos^{-1}(\frac{x}{a})$Differentiating with respect to $x$:$\frac{dy}{dx} = \frac{1}{2} \cdot \left( -\frac{1}{\sqrt{1 - (\frac{x}{a})^2}} ight) \cdot \frac{1}{a}$$\frac{dy}{dx} = -\frac{1}{2} \cdot \frac{1}{\sqrt{\frac{a^2 - x^2}{a^2}}} \cdot \frac{1}{a} = -\frac{1}{2} \cdot \frac{a}{\sqrt{a^2 - x^2}} \cdot \frac{1}{a} = -\frac{1}{2\sqrt{a^2 - x^2}}$.

If $y = \tan^{-1} \sqrt{\frac{a - x}{a + x}}$,then $\frac{dy}{dx} = $

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