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

Answer

Suppose that then$
a =\tan \theta_1, b=\tan \theta_2$
$\theta_1 =\tan ^{-1} a, \theta_2=\tan ^{-1} b$
$\therefore \frac{1-a^2}{1+a^2}=\frac{1-\tan ^2 \theta_1}{1+\tan ^2 \theta_1}=\cos 2 \theta_1$
and $ \frac{1-b^2}{1+b^2}=\frac{1-\tan ^2 \theta_2}{1+\tan ^2 \theta_2}=\cos 2 \theta_2$
hence from given equation :$\cos ^{-1}\left(\cos 2 \theta_1\right)+\cos ^{-1}\left(\cos 2 \theta_2\right)=2 \tan ^{-1} x$
$2 \theta_1+2 \theta_2=2 \tan ^{-1} x$
$\theta_1+\theta_2=\tan ^{-1} x$
$\Rightarrow \tan \left(\theta_1+\theta_2\right)=x$
$\Rightarrow \frac{\tan \theta_1+\tan \theta_2}{1-\tan \theta_1 \tan \theta_2}=x$
$\Rightarrow \frac{a+b}{1-a b}=x$
$\therefore x=\frac{a+b}{1-a b} \text {}$

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