Evaluate: tan^{-1} left( frac{1 - x^2}{2x} ri | vedclass

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(B) Let $x = 	an 	heta$. Then $	heta = 	an^{-1} x$.The given expression is $	an^{-1} \left( \frac{1 - x^2}{2x} ight) + \cos^{-1} \left( \frac{1

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$\frac{\pi}{2}$

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(B) Let $x = 	an 	heta$. Then $	heta = 	an^{-1} x$.The given expression is $	an^{-1} \left( \frac{1 - x^2}{2x} ight) + \cos^{-1} \left( \frac{1 - x^2}{1 + x^2} ight)$.Substitute $x = 	an 	heta$:$= 	an^{-1} \left( \frac{1 - 	an^2 	heta}{2 	an 	heta} ight) + \cos^{-1} \left( \frac{1 - 	an^2 	heta}{1 + 	an^2 	heta} ight)$Using trigonometric identities $\cot 2	heta = \frac{1 - 	an^2 	heta}{2 	an 	heta}$ and $\cos 2	heta = \frac{1 - 	an^2 	heta}{1 + 	an^2 	heta}$:$= 	an^{-1} (\cot 2	heta) + \cos^{-1} (\cos 2	heta)$Since $	an^{-1} (\cot 2	heta) = 	an^{-1} \left( 	an \left( \frac{\pi}{2} - 2	heta ight) ight) = \frac{\pi}{2} - 2	heta$ and $\cos^{-1} (\cos 2	heta) = 2	heta$:$= \left( \frac{\pi}{2} - 2	heta ight) + 2	heta$$= \frac{\pi}{2}$.

Evaluate: $\tan^{-1} \left( \frac{1 - x^2}{2x} \right) + \cos^{-1} \left( \frac{1 - x^2}{1 + x^2} \right)$

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