The value of tan ^{-1}left(frac{sqrt{1+x^2}+s | vedclass

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(A) Let $T = 	an ^{-1}\left(\frac{\sqrt{1+x^2}+\sqrt{1-x^2}}{\sqrt{1+x^2}-\sqrt{1-x^2}}ight)$.Put $x^2 = \cos 2	heta$,which implies $2	heta = \co

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

Vedclass · Answer

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

The value of $\tan ^{-1}\left(\frac{\sqrt{1+x^2}+\sqrt{1-x^2}}{\sqrt{1+x^2}-\sqrt{1-x^2}}\right)$ for $|x| < \frac{1}{\sqrt{2}}, x \neq 0$ is:

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