The value of $ an ^{-1}\left(\frac{\sqrt{1+x^2}+\sqrt{1-x^2}}{\sqrt{1+x^2}-\sqrt{1-x^2}} ight)$, $|x|

Question

The value of $	an ^{-1}\left(\frac{\sqrt{1+x^2}+\sqrt{1-x^2}}{\sqrt{1+x^2}-\sqrt{1-x^2}}ight)$, $|x| < \frac{1}{2}, x 
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Aicharya · Accepted Answer

Let $\mathrm{T}=	an ^{-1}\left(\frac{\sqrt{1+x^2}+\sqrt{1-x^2}}{\sqrt{1+x^2}-\sqrt{1-x^2}}ight)$ Put $x^2=\cos 2 	heta \Rightarrow 	heta=\frac{1}{2} \cos ^{-1} x^2$ $\begin{aligned} 	herefore \quad \mathrm{T} & =	an ^{-1}\left(\frac{\sqrt{1+\cos 2 	heta}+\sqrt{1-\cos 2 	heta}}{\sqrt{1+\cos 2 	heta}-\sqrt{1-\cos 2 	heta}}ight) \ & =	an ^{-1}\left(\frac{\sqrt{2} \cos 	heta+\sqrt{2} \sin 	heta}{\sqrt{2} \cos 	heta-\sqrt{2} \sin 	heta}ight) \ & =	an ^{-1}\left(\frac{\cos 	heta+\sin 	heta}{\cos 	heta-\sin 	heta}ight) \ & =	an ^{-1}\left(\frac{1+	an 	heta}{1-	an 	heta}ight) \ & =	an ^{-1}\left(	an \left(\frac{\pi}{4}+	hetaight)ight) \ & =\frac{\pi}{4}+	heta \ & =\frac{\pi}{4}+\frac{1}{2} \cos ^{-1} x^2 \end{aligned}$

The value of $\tan ^{-1}\left(\frac{\sqrt{1+x^2}+\sqrt{1-x^2}}{\sqrt{1+x^2}-\sqrt{1-x^2}}\right)$, $|x| <…

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