Question

Prove that:
${\cos }^{-1}\frac{x-x^{-1}}{x+x^{-1}}=2{\tan }^{-1}\frac{1}{x}$

Answer 1

$LHS={\text{cos}}^{-1}(\frac{x-x^{-1}}{x+x^{-1}})={\text{cos}}^{-1}(\frac{1-\frac{1}{x^2}}{1+\frac{1}{x^2}})$

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

$⟹LHS={\text{cos}}^{-1}(\frac{1-{\tan }^2\theta }{1+{\tan }^2\theta })={\text{cos}}^{-1}\left(\cos 2\theta \right)=2\theta =2{\text{tan}}^{-1}(\frac{1}{x})$