Let f(x) = cos^{-1}left(frac{2x}{1+x^2}right) | vedclass

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(D) We know that $\cos^{-1}(y) = \frac{\pi}{2} - \sin^{-1}(y)$ and $\sin^{-1}(y) = \frac{\pi}{2} - \cos^{-1}(y)$.Given $f(x) = \cos^{-1}\left(\frac{2x

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$2\pi$

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(D) We know that $\cos^{-1}(y) = \frac{\pi}{2} - \sin^{-1}(y)$ and $\sin^{-1}(y) = \frac{\pi}{2} - \cos^{-1}(y)$.Given $f(x) = \cos^{-1}\left(\frac{2x}{1+x^2}ight) + \sin^{-1}\left(\frac{1-x^2}{1+x^2}ight)$.Using the substitution $x = 	an 	heta$,we have $\frac{2x}{1+x^2} = \sin(2	heta)$ and $\frac{1-x^2}{1+x^2} = \cos(2	heta)$.For $x \ge 1$,$f(x) = \left(\frac{\pi}{2} - \sin^{-1}\left(\frac{2x}{1+x^2}ight)ight) + \left(\frac{\pi}{2} - \cos^{-1}\left(\frac{1-x^2}{1+x^2}ight)ight)$.For $x \ge 1$,$\sin^{-1}\left(\frac{2x}{1+x^2}ight) = 2	an^{-1}(x) - \pi$ is not correct; rather,for $x > 1$,$\sin^{-1}\left(\frac{2x}{1+x^2}ight) = \pi - 2	an^{-1}(x)$ and $\cos^{-1}\left(\frac{1-x^2}{1+x^2}ight) = 2	an^{-1}(x)$.Thus,$f(x) = (\pi - 2	an^{-1}(x)) + (2	an^{-1}(x)) = \pi$ for $x \ge 1$.Therefore,$f(1) = \pi$ and $f(2) = \pi$.Hence,$f(1) + f(2) = \pi + \pi = 2\pi$.

Let $f(x) = \cos^{-1}\left(\frac{2x}{1+x^2}\right) + \sin^{-1}\left(\frac{1-x^2}{1+x^2}\right)$,then the value of $f(1) + f(2)$ is -

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