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If $\sin ^{-1}\left(\frac{2 \mathrm{a}}{1+\mathrm{a}^{2}}\right)+\sin ^{-1}\left(\frac{2 \mathrm{~b}}{1+\mathrm{b}^{2}}\right)=2 \tan ^{-1} \mathrm{x}$, then $\mathrm{x}$ is equal to
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Verified Answer
The correct answer is:
$\frac{a+b}{1-a b}$
Since a $>0, b>0$ and $2 \tan ^{-1} x=\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)$
$\therefore$ Given expression is $2 \tan ^{-1} \mathrm{a}+2 \tan ^{-1} \mathrm{~b}=2 \tan ^{-1} \mathrm{x}$
$\Rightarrow 2 \tan ^{-1}\left(\frac{a+b}{1-a b}\right)=2 \tan ^{-1} x$
$\Rightarrow x=\frac{a+b}{1-a b}$
$\therefore$ Given expression is $2 \tan ^{-1} \mathrm{a}+2 \tan ^{-1} \mathrm{~b}=2 \tan ^{-1} \mathrm{x}$
$\Rightarrow 2 \tan ^{-1}\left(\frac{a+b}{1-a b}\right)=2 \tan ^{-1} x$
$\Rightarrow x=\frac{a+b}{1-a b}$
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