Question

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

Answer 1

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

LHS= $2\tan^{-1}\frac {1}{2}+\tan^{-1}\frac{1}{7}$

= $\tan^{-1} \frac{\frac{2.1}{2}}{1-(\frac{1}{2})^2}+\tan^{-1}\frac{1}{7} \ [2tan^{-1x}=\tan^{-1}\frac{2x}{1-x^2}]$

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

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

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

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

= $\tan^{-1}\frac{31}{17}$=RHS