Question

If value of ${\tan }^{-1}\frac{1}{5}+{\tan }^{-1}\frac{1}{8}={\tan }^{-1}\frac{1}{m}$.Find $m$

Answer 1

$\tan \frac{1}{5}+{\tan }^{-1}\frac{1}{8}={\tan }^{-1}\frac{1}{m}$

$\tan \frac{\frac{1}{5}+\frac{1}{8}}{1-(\frac{1}{5}\times \frac{1}{8})}={\tan }^{-1}\frac{1}{m}\{\because {\tan }^{-1}x+{\tan }^{-1}y={\tan }^{-1}\frac{x+y}{1-xy}\}$

${\tan }^{-1}\frac{\frac{13}{40}}{1-\frac{1}{40}}={\tan }^{-1}\frac{1}{m}$

${\tan }^{-1}\frac{1}{3}={\tan }^{-1}\frac{1}{m}$

$\Rightarrow \frac{1}{3}=\frac{1}{m}$

$\Rightarrow m=3$