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

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(A) Let $x^2 = \cos 2	heta$,which implies $	heta = \frac{1}{2} \cos^{-1} x^2$.Substituting this into the expression:$	an^{-1} \left[ \frac{\sqrt{1+

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$\frac{\pi}{4} + \frac{1}{2} \cos^{-1} x^2$

Vedclass · Answer

(A) Let $x^2 = \cos 2	heta$,which implies $	heta = \frac{1}{2} \cos^{-1} x^2$.Substituting this into the expression:$	an^{-1} \left[ \frac{\sqrt{1+\cos 2	heta} + \sqrt{1-\cos 2	heta}}{\sqrt{1+\cos 2	heta} - \sqrt{1-\cos 2	heta}} ight]$Using the identities $1+\cos 2	heta = 2\cos^2 	heta$ and $1-\cos 2	heta = 2\sin^2 	heta$:$= 	an^{-1} \left[ \frac{\sqrt{2}\cos 	heta + \sqrt{2}\sin 	heta}{\sqrt{2}\cos 	heta - \sqrt{2}\sin 	heta} ight]$Dividing numerator and denominator by $\sqrt{2}\cos 	heta$:$= 	an^{-1} \left[ \frac{1 + 	an 	heta}{1 - 	an 	heta} ight]$Using the formula $	an(\frac{\pi}{4} + 	heta) = \frac{	an \frac{\pi}{4} + 	an 	heta}{1 - 	an \frac{\pi}{4} 	an 	heta}$:$= 	an^{-1} \left[ 	an \left( \frac{\pi}{4} + 	heta ight) ight] = \frac{\pi}{4} + 	heta$Substituting back $	heta = \frac{1}{2} \cos^{-1} x^2$:$= \frac{\pi}{4} + \frac{1}{2} \cos^{-1} x^2$.

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

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