cot ^{-1}left(frac{sqrt{1+x^2}-1}{x}right) = | vedclass

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Question

(C) Let $x = 	an 	heta$,where $	heta = 	an^{-1} x$. Then,$\sqrt{1+x^2} = \sqrt{1+	an^2 	heta} = \sqrt{\sec^2 	heta} = \sec 	heta$. The express

Vedclass · Accepted Answer

$\frac{\pi}{2}-\frac{1}{2} 	an ^{-1} x$

Vedclass · Answer

(C) Let $x = 	an 	heta$,where $	heta = 	an^{-1} x$. Then,$\sqrt{1+x^2} = \sqrt{1+	an^2 	heta} = \sqrt{\sec^2 	heta} = \sec 	heta$. The expression becomes $\cot^{-1}\left(\frac{\sec 	heta - 1}{	an 	heta}ight)$. $= \cot^{-1}\left(\frac{\frac{1}{\cos 	heta} - 1}{\frac{\sin 	heta}{\cos 	heta}}ight) = \cot^{-1}\left(\frac{1 - \cos 	heta}{\sin 	heta}ight)$. Using half-angle formulas,$1 - \cos 	heta = 2 \sin^2 \frac{	heta}{2}$ and $\sin 	heta = 2 \sin \frac{	heta}{2} \cos \frac{	heta}{2}$. $= \cot^{-1}\left(\frac{2 \sin^2 \frac{	heta}{2}}{2 \sin \frac{	heta}{2} \cos \frac{	heta}{2}}ight) = \cot^{-1}\left(	an \frac{	heta}{2}ight)$. Since $\cot^{-1}(	an \alpha) = \frac{\pi}{2} - 	an^{-1}(	an \alpha) = \frac{\pi}{2} - \alpha$,we have: $= \frac{\pi}{2} - \frac{	heta}{2} = \frac{\pi}{2} - \frac{1}{2} 	an^{-1} x$.

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

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