if the

Question:

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) $\frac{2 a}{1-a^{2}}$

Solution:

$\sin ^{-1}\left(\frac{2 a}{1-a^{2}}\right)+\cos ^{-1}\left(\frac{1-a^{2}}{1+a^{2}}\right)=\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)$

$\Rightarrow 2 \tan ^{-1} a+2 \tan ^{-1} a=2 \tan ^{-1} x$

$\Rightarrow 4 \tan ^{-1} a=2 \tan ^{-1} x$

$\Rightarrow 2 \tan ^{-1} a=\tan ^{-1} x$

$\Rightarrow \tan ^{-1}\left(\frac{2 a}{1-a^{2}}\right)=\tan ^{-1} x$

$\Rightarrow x=\frac{2 a}{1-a^{2}}$

Hence, the correct answer is option(d).

 

 

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