If $\mathrm{y}= an ^{-1}\left[\frac{x-\sqrt{1-x^{2}}}{x+\sqrt{1-x^{2}}} ight]$, then $\left(\frac{d y}{d x} ight)=$

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If $\mathrm{y}=	an ^{-1}\left[\frac{x-\sqrt{1-x^{2}}}{x+\sqrt{1-x^{2}}}ight]$, then $\left(\frac{d y}{d x}ight)=$

Aicharya · Accepted Answer

Given $y=	an ^{-1}\left[\frac{x-\sqrt{1-x^{2}}}{x+\sqrt{1-x^{2}}}ight]$ Put $x=\cos 	heta \Rightarrow 	heta=\cos ^{-1} x$ $y=	an ^{-1}\left[\frac{\cos 	heta-\sin 	heta}{\cos 	heta+\sin 	heta}ight]=	an ^{-1}\left[\frac{1-	an 	heta}{1+	an 	heta}ight]$ $\quad=	an ^{-1}\left[	an \left(\frac{\pi}{4}-	hetaight)ight]=\frac{\pi}{4}-	heta$ $	herefore \frac{d y}{d x}=\frac{\pi}{4}-\cos ^{-1} x$ $\begin{aligned} y &=\frac{1}{\sqrt{1-x^{2}}} \ y &=\frac{1}{\sqrt{1-x^{2}}} \end{aligned}$

If $\mathrm{y}=\tan ^{-1}\left[\frac{x-\sqrt{1-x^{2}}}{x+\sqrt{1-x^{2}}}\right]$, then $\left(\frac{d y}{d…

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