Question
Prove that $\tan ^{-1} \frac{2}{3}=\frac{1}{2} \tan ^{-1} \frac{12}{5}$

Answer

$\text{L.H.S}. =\tan ^{-1} \frac{2}{3}= A \ ($suppose$)$
or $\tan A=\frac{2}{3}$
$\tan 2 A=\frac{2 \tan A}{1-\tan ^2 A}$
$=\frac{2 \times \frac{2}{3}}{1-\left(\frac{2}{3}\right)^2}$
$=\frac{\frac{4}{3}}{1-\frac{4}{9}}=\frac{\frac{4}{3}}{\frac{5}{9}}$
$\Rightarrow \tan 2 A=\frac{4}{3} \times \frac{9}{5}=\frac{12}{5}$
$\Rightarrow 2 A=\tan ^{-1}\left(\frac{12}{5}\right)$
$\Rightarrow A =\frac{1}{2} \tan ^{-1}\left(\frac{12}{5}\right)$
$\Rightarrow \tan ^{-1}\left(\frac{2}{3}\right)$
$=\frac{1}{2} \tan ^{-1}\left(\frac{12}{5}\right) $ Hence proved.

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