Show that $2 \tan ^{-1} x+\sin ^{-1} \frac{2 x}{1+x^{2}}$ is constant for $x \geq 1$, find that constant.
We have
$2 \tan ^{-1} x+\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)$
(1) For $x>1$,
$=2 \tan ^{-1} x+\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)$
$=\pi-\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)+\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right) \quad\left[\because 2 \tan ^{-1} x=\pi-\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right), x>1\right]$
$=\pi$
(2) For $x=1$
$=2 \tan ^{-1} x+\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)$
$=2 \tan ^{-1}(1)+\sin ^{-1}\left(\frac{2(1)}{1+(1)^{2}}\right)$
$=2 \tan ^{-1}(1)+\sin ^{-1}(1)$
$=2\left(\frac{\pi}{4}\right)+\frac{\pi}{2}$
$=\pi$
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