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

Answer

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

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