Prove
Question:

Prove $2 \tan ^{-1} \frac{1}{2}+\tan ^{-1} \frac{1}{7}=\tan ^{-1} \frac{31}{17}$

Solution:

To prove: $2 \tan ^{-1} \frac{1}{2}+\tan ^{-1} \frac{1}{7}=\tan ^{-1} \frac{31}{17}$

L.H.S. $=2 \tan ^{-1} \frac{1}{2}+\tan ^{-1} \frac{1}{7}$

$=\tan ^{-1} \frac{2 \cdot \frac{1}{2}}{1-\left(\frac{1}{2}\right)^{2}}+\tan ^{-1} \frac{1}{7} \quad\left[2 \tan ^{-1} x=\tan ^{-1} \frac{2 x}{1-x^{2}}\right]$

$=\tan ^{-1} \frac{1}{\left(\frac{3}{4}\right)}+\tan ^{-1} \frac{1}{7}$

$=\tan ^{-1} \frac{4}{3}+\tan ^{-1} \frac{1}{7}$

$=\tan ^{-1} \frac{\frac{4}{3}+\frac{1}{7}}{1-\frac{4}{3} \cdot \frac{1}{7}}$    $\left[\tan ^{-1} x+\tan ^{-1} y=\tan ^{-1} \frac{x+y}{1-x y}\right]$

$=\tan ^{-1} \frac{\left(\frac{28+3}{21}\right)}{\left(\frac{21-4}{21}\right)}$

$=\tan ^{-1} \frac{31}{17}=$ R.H.S.

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