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$2 \tan ^{-1} \frac{1}{2}+\tan ^{-1} \frac{1}{7}=\tan ^{-1} \frac{31}{17}$
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Verified Answer
L.H.S. $2 \tan ^{-1} \frac{1}{2}+\tan ^{-1} \frac{1}{7}$
$$
\begin{aligned}
&=\tan ^{-1} \frac{2 \times \frac{1}{2}}{1-\frac{1}{4}}+\tan ^{-1} \frac{1}{7}=\tan ^{-1} \frac{4}{3}+\tan ^{-1} \frac{1}{7} \\
&=\tan ^{-1} \frac{\frac{4}{3}+\frac{1}{7}}{1-\frac{4}{3} \times \frac{1}{7}}=\tan ^{-1}\left(\frac{28+3}{21-4}\right)=\tan ^{-1} \frac{31}{17}
\end{aligned}
$$
$$
\begin{aligned}
&=\tan ^{-1} \frac{2 \times \frac{1}{2}}{1-\frac{1}{4}}+\tan ^{-1} \frac{1}{7}=\tan ^{-1} \frac{4}{3}+\tan ^{-1} \frac{1}{7} \\
&=\tan ^{-1} \frac{\frac{4}{3}+\frac{1}{7}}{1-\frac{4}{3} \times \frac{1}{7}}=\tan ^{-1}\left(\frac{28+3}{21-4}\right)=\tan ^{-1} \frac{31}{17}
\end{aligned}
$$
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