The value of sin left[tan ^{-1}left(frac{1-x^ | vedclass

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(B) Let $x = 	an 	heta$. Then $	heta = 	an^{-1} x$. Substituting this into the expression: $	an^{-1}\left(\frac{1-	an^2 	heta}{2 	an 	heta}

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$1$

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(B) Let $x = 	an 	heta$. Then $	heta = 	an^{-1} x$. Substituting this into the expression: $	an^{-1}\left(\frac{1-	an^2 	heta}{2 	an 	heta}ight) = 	an^{-1}\left(\frac{1}{	an 2	heta}ight) = 	an^{-1}(\cot 2	heta) = 	an^{-1}\left(	an\left(\frac{\pi}{2} - 2	hetaight)ight) = \frac{\pi}{2} - 2	heta$. Next,$\cos^{-1}\left(\frac{1-	an^2 	heta}{1+	an^2 	heta}ight) = \cos^{-1}(\cos 2	heta) = 2	heta$. Adding these two parts: $(\frac{\pi}{2} - 2	heta) + 2	heta = \frac{\pi}{2}$. Finally,$\sin\left(\frac{\pi}{2}ight) = 1$.

The value of $\sin \left[\tan ^{-1}\left(\frac{1-x^2}{2 x}\right)+\cos ^{-1}\left(\frac{1-x^2}{1+x^2}\right)\right]$ is

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