If $\sin ^{-1}\left(\frac{2 a}{1-a^{2}}\right)+\cos ^{-1}\left(\frac{1-a^{2}}{1+a^{2}}\right)=\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)$, where $a, x \in(0,1)$, then, the value of $x$ is
(a) 0
(b) $\frac{a}{2}$
(c) $a$
(d) $\frac{2 a}{1-a^{2}}$
$\sin ^{-1}\left(\frac{2 a}{1-a^{2}}\right)+\cos ^{-1}\left(\frac{1-a^{2}}{1+a^{2}}\right)=\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)$
$\Rightarrow 2 \tan ^{-1} a+2 \tan ^{-1} a=2 \tan ^{-1} x$
$\Rightarrow 4 \tan ^{-1} a=2 \tan ^{-1} x$
$\Rightarrow 2 \tan ^{-1} a=\tan ^{-1} x$
$\Rightarrow \tan ^{-1}\left(\frac{2 a}{1-a^{2}}\right)=\tan ^{-1} x$
$\Rightarrow x=\frac{2 a}{1-a^{2}}$
Hence, the correct answer is option(d).
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.