Question

Show that $2{\tan }^{-1}\frac{1}{2}+{\tan }^{-1}\frac{1}{7}={\tan }^{-1}\frac{31}{17}$.

Answer 1

We know that:

$2{\tan }^{-1}x={\tan }^{-1}\frac{2x}{1-x^2}$

$2{\tan }^{-1}\frac{1}{2}={\tan }^{-1}\frac{2\times \frac{1}{2}}{1-{\left(\frac{1}{2}\right)}^2}={\tan }^{-1}\frac{4}{3}$

Now, using the formula:

${\tan }^{-1}A+{\tan }^{-1}B={\tan }^{-1}\frac{A+B}{1-AB}$

${\tan }^{-1}\frac{4}{3}+{\tan }^{-1}\frac{1}{7}={\tan }^{-1}\frac{\frac{4}{3}+\frac{1}{7}}{1-\frac{4}{3}\times \frac{1}{7}}={\tan }^{-1}\frac{31}{17}$