Question

Solve: ${\displaystyle {\tan }^{-1}x={\cos }^{-1} \frac{1-{\text{a}}^2}{1+{\text{a}}^2}-{\cos }^{-1} \frac{1-{\text{b}}^2}{1+{\text{b}}^2},\text{a}>0,\text{b}>0}$

Answer 1

${\displaystyle 2{\tan }^{-1}x={\cos }^{-1}\left(\frac{1-{\text{a}}^2}{1+{\text{a}}^2}\right)-{\cos }^{-1}\left(\frac{1-{\text{b}}^2}{1+{\text{b}}^2}\right)\text{a}>0}$

W.K.T ${\displaystyle 2{\tan }^{-1}x={\cos }^{-1}\left(\frac{1-x^2}{1+x^2}\right)}$

${\displaystyle {\cos }^{-1}\left[\frac{1-{\text{a}}^2}{1+{\text{a}}^2}\right]=2{\tan }^{-1}\text{a}}$

${\displaystyle {\cos }^{-1}\left[\frac{1-{\text{b}}^2}{1+{\text{b}}^2}\right]=2{\tan }^{-1}\text{b}}$

R.H.S ${\displaystyle {\cos }^{-1}\left[\frac{1-{\text{a}}^2}{1+{\text{a}}^2}\right]-{\cos }^{-1}\left[\frac{1-{\text{b}}^2}{1+{\text{b}}^2}\right]}$

= ${\displaystyle 2{\tan }^{-1}\text{a}-2ta{m}^{-1}\text{b}}$

= ${\displaystyle 2[{\tan }^{-1}\text{a}-{\tan }^{-1}\text{b}]}$

= ${\displaystyle 2\left[\frac{{\tan }^{-1}(\text{a}-\text{b})}{1+\text{ab}}\right]}$

L.H.S = R.H.S

${\displaystyle 2{\tan }^1\left[\frac{\text{a}-\text{b}}{1+\text{ab}}\right]=2{\tan }^{-x}}$

x = ${\displaystyle \frac{\text{a}-\text{b}}{1+\text{ab}},\text{a}>0}$