If

Question:

If $\sin ^{-1} \frac{2 a}{1+a^{2}}+\sin ^{-1} \frac{2 b}{1+b^{2}}=2 \tan ^{-1} x$, prove that $x=\frac{a+b}{1-a b}$

Solution:

Let:

$a=\tan z$

$b=\tan y$

Then,

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

$\Rightarrow \sin ^{-1} \frac{2 \tan z}{1+\tan ^{2} z}+\sin ^{-1} \frac{2 \tan y}{1+\tan ^{2} y}=2 \tan ^{-1} x$

$\Rightarrow \sin ^{-1}(\sin 2 z)+\sin ^{-1}(\sin 2 y)=2 \tan ^{-1} x$                   $\left[\because \sin 2 x=\frac{2 \tan x}{1+\tan ^{2} x}\right]$

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

$\Rightarrow \tan ^{-1} a+\tan ^{-1} b=\tan ^{-1} x \quad[\because a=\tan z$ and $b=\tan y]$

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

$\Rightarrow x=\frac{a+b}{1-a b}$

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