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

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$f(\mathrm{x})=\frac{\pi}{2}-\sin ^{-1}\left(\frac{2 \mathrm{x}}{1+\mathrm{x}^{2}}\right)+\frac{\pi}{2}-\cos ^{-1}\left(\frac{1-\mathrm{x}^{2}}{1+\mat

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$f(\mathrm{x})=\frac{\pi}{2}-\sin ^{-1}\left(\frac{2 \mathrm{x}}{1+\mathrm{x}^{2}}\right)+\frac{\pi}{2}-\cos ^{-1}\left(\frac{1-\mathrm{x}^{2}}{1+\mathrm{x}^{2}}\right)$

$=\pi-\left(\sin ^{-1}\left(\frac{2 \mathrm{x}}{1+\mathrm{x}^{2}}\right)+\cos ^{-1}\left(\frac{1-\mathrm{x}^{2}}{1+\mathrm{x}^{2}}\right)\right)$

$f(1)=\pi-\left(\sin ^{-1}(1)+\cos ^{-1}(0)\right)=0$

$f(2)=\pi-\left(\sin ^{-1}\left(\frac{4}{5}\right)+\cos ^{-1}\left(\frac{-3}{5}\right)\right)=0$

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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