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Answer: $\sin ^{-1} \frac{2 a}{1+a^{2}}+\sin ^{-1} \frac{2 b}{1+b^{2}}=2 \tan ^{-1} x$to prove $x=\frac{a+b}{1-a b}$

Hints: First we will convert L.H.S of the question in $\tan ^{-1}$

Given: $\sin ^{-1} \frac{2 a}{1+a^{2}}+\sin ^{-1} \frac{2 b}{1+b^{2}}=2 \tan ^{-1} x$

Explanation:

\begin{aligned} &\sin ^{-1} \frac{2 a}{1+a^{2}}+\sin ^{-1} \frac{2 b}{1+b^{2}}=2 \tan ^{-1} x \\ &\because 2 \tan ^{-1} a=\sin ^{-1}\left(\frac{2 a}{1+a^{2}}\right) \\ &\therefore \sin ^{-1} \frac{2 a}{1+a^{2}}+\sin ^{-1} \frac{2 b}{1+b^{2}}=2 \tan ^{-1} x \end{aligned}

\begin{aligned} &2 \tan ^{-1} a+2 \tan ^{-1} b=2 \tan ^{-1} x \\ &2\left(\tan ^{-1} a+\tan ^{-1} b\right)=2 \tan ^{-1} x \\ &\tan ^{-1} a+\tan ^{-1} b=\tan ^{-1} x \end{aligned}

$\tan ^{-1}\left(\frac{a+b}{1-a b}\right)=\tan ^{-1} x$                                                                    $\left[\because \tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)\right]$

\begin{aligned} &\frac{a+b}{1-a b}=\tan \left(\tan ^{-1} x\right) \\ &\frac{a+b}{1-a b}=x \end{aligned}

Hence it is proved that $x=\frac{a+b}{1-a b}$

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