tan^{-1} left( frac{sqrt{1 + x^2} - 1}{x} rig | vedclass

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(B) Let $x = 	an 	heta$,then $	heta = 	an^{-1} x$.Substituting this into the expression:$	an^{-1} \left( \frac{\sqrt{1 + 	an^2 	heta} - 1}{	an

Vedclass · Accepted Answer

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

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(B) Let $x = 	an 	heta$,then $	heta = 	an^{-1} x$.Substituting this into the expression:$	an^{-1} \left( \frac{\sqrt{1 + 	an^2 	heta} - 1}{	an 	heta} ight) = 	an^{-1} \left( \frac{\sec 	heta - 1}{	an 	heta} ight)$$= 	an^{-1} \left( \frac{\frac{1}{\cos 	heta} - 1}{\frac{\sin 	heta}{\cos 	heta}} ight) = 	an^{-1} \left( \frac{1 - \cos 	heta}{\sin 	heta} ight)$Using the half-angle formulas $1 - \cos 	heta = 2 \sin^2 \frac{	heta}{2}$ and $\sin 	heta = 2 \sin \frac{	heta}{2} \cos \frac{	heta}{2}$:$= 	an^{-1} \left( \frac{2 \sin^2 \frac{	heta}{2}}{2 \sin \frac{	heta}{2} \cos \frac{	heta}{2}} ight) = 	an^{-1} \left( 	an \frac{	heta}{2} ight)$$= \frac{	heta}{2} = \frac{1}{2} 	an^{-1} x$.

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

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