Question

Simplify: ${\tan }^{-1}\frac{x-1}{x+1}$

Answer 1

Let $x=\tan \theta ⟹\theta ={\text{tan}}^{-1}x$

${\text{tan}}^{-1}\frac{x-1}{x+1}={\text{tan}}^{-1}\frac{\tan \theta -1}{\tan \theta +1}={\text{tan}}^{-1}\frac{\tan \theta -\tan \left(\pi /4\right)}{1+\tan \theta \tan \left(\pi /4\right)}={\text{tan}}^{-1}\left(\tan \right(\theta -\pi /4\left)\right)=\theta -\pi /4={\text{tan}}^{-1}x-\frac{\pi }{4}$