Prove that $\tan \left[\frac{1}{2} \sin ^{-1}\left(\frac{2 x}{1+x^2}\right)+\frac{1}{2} \cos ^{-1}\left(\frac{1-x^2}{1+x^2}\right)\right]=\frac{2 x}{1-x^2}$
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Prove that $\tan \left[\frac{1}{2} \sin ^{-1}\left(\frac{2 x}{1+x^2}\right)+\frac{1}{2} \cos ^{-1}\left(\frac{1-x^2}{1+x^2}\right)\right]=\frac{2 x}{1-x^2}$

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$\text { L.H.S. }  =\tan \left[\frac{1}{2} \sin ^{-1}\left(\frac{2 x}{1+x^2}\right)+\frac{1}{2} \cos ^{-1}\left(\frac{1-x^2}{1+x^2}\right)\right]$

$=\tan \left[\frac{1}{2} \cdot 2 \tan ^{-1}(x)+\frac{1}{2} \times 2 \tan ^{-1}(x)\right]$

$=\tan \left[2 \tan ^{-1}(x)\right]=\tan \left[\tan ^{-1} \frac{2 x}{1-x^2}\right]=\frac{2 x}{1-x^2}=\text { R.H.S. }$

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