Question

Prove that $tan[ta{n}^{-1}\frac{m}{n}-ta{n}^{-1}\left(\frac{m-n}{m+n}\right)]=1$

Answer 1

Let $\tan x=\frac{m}{n}⟹x={\text{tan}}^{-1}\frac{m}{n}$

${\text{tan}}^{-1}(\frac{m-n}{m+n})={\text{tan}}^{-1}(\frac{\frac{m}{n}-1}{\frac{m}{n}+1})={\text{tan}}^{-1}(\frac{\tan x-\tan \pi /4}{1+\tan x\tan \pi /4})={\text{tan}}^{-1}(\tan (x-\frac{\pi }{4}))=x-\frac{\pi }{4}={\text{tan}}^{-1}\frac{m}{n}-\frac{\pi }{4}$

$\tan [{\text{tan}}^{-1}\frac{m}{n}-{\text{tan}}^{-1}\left(\frac{m-n}{m+n}\right)]=\tan [{\text{tan}}^{-1}\frac{m}{n}-{\text{tan}}^{-1}\frac{m}{n}+\frac{\pi }{4}]=\tan \frac{\pi }{4}=1$