If A=2 tan ^{-1}left(frac{1+x}{1-x}right) and | vedclass

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(D) Let $x = 	an 	heta$. Since $x \in (0, 1)$,we have $	heta \in (0, \frac{\pi}{4})$.First,consider $A = 2 	an^{-1}\left(\frac{1+x}{1-x}ight)$.U

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$\frac{\pi}{2}$

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(D) Let $x = 	an 	heta$. Since $x \in (0, 1)$,we have $	heta \in (0, \frac{\pi}{4})$.First,consider $A = 2 	an^{-1}\left(\frac{1+x}{1-x}ight)$.Using the formula $	an^{-1}\left(\frac{1+	an 	heta}{1-	an 	heta}ight) = 	an^{-1}(	an(\frac{\pi}{4} + 	heta)) = \frac{\pi}{4} + 	heta$,we get $A = 2(\frac{\pi}{4} + 	heta) = \frac{\pi}{2} + 2	heta$.Next,consider $B = \cos^{-1}\left(\frac{1-x^2}{1+x^2}ight)$.Using the formula $\cos^{-1}\left(\frac{1-	an^2 	heta}{1+	an^2 	heta}ight) = \cos^{-1}(\cos 2	heta) = 2	heta$.Finally,$A - B = (\frac{\pi}{2} + 2	heta) - 2	heta = \frac{\pi}{2}$.

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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