If $A=2 an ^{-1}\left(\frac{1+x}{1-x} ight)$ and $B=\cos ^{-1}\left(\frac{1-x^2}{1+x^2} ight)$, where $x \in(0,1)$, then $A-B=$

Question

If $A=2 	an ^{-1}\left(\frac{1+x}{1-x}ight)$ and $B=\cos ^{-1}\left(\frac{1-x^2}{1+x^2}ight)$, where $x \in(0,1)$, then $A-B=$

Aicharya · Accepted Answer

Let $x=	an 	heta$ then $\begin{aligned} & A=2 	an ^{-1}\left(	an \left(\frac{\pi}{4}+	hetaight)ight) 	ext { and } B=\cos ^{-1}(\cos 2 	heta) \ & \Rightarrow A=2\left\{\frac{\pi}{4}+	hetaight\} 	ext { and } B=2 	heta \ & \Rightarrow A-B=\frac{\pi}{2}\end{aligned}$

If $A=2 \tan ^{-1}\left(\frac{1+x}{1-x}\right)$ and $B=\cos ^{-1}\left(\frac{1-x^2}{1+x^2}\right)$, where $x…

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