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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 $x \in(0,1)$, then $A-B=$
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The correct answer is:
$\frac{\pi}{2}$
Let $x=\tan \theta$ then
$\begin{aligned} & A=2 \tan ^{-1}\left(\tan \left(\frac{\pi}{4}+\theta\right)\right) \text { and } B=\cos ^{-1}(\cos 2 \theta) \\ & \Rightarrow A=2\left\{\frac{\pi}{4}+\theta\right\} \text { and } B=2 \theta \\ & \Rightarrow A-B=\frac{\pi}{2}\end{aligned}$
$\begin{aligned} & A=2 \tan ^{-1}\left(\tan \left(\frac{\pi}{4}+\theta\right)\right) \text { and } B=\cos ^{-1}(\cos 2 \theta) \\ & \Rightarrow A=2\left\{\frac{\pi}{4}+\theta\right\} \text { and } B=2 \theta \\ & \Rightarrow A-B=\frac{\pi}{2}\end{aligned}$
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