Prove that `2sin^-1[3/5]-tan^-1[17/31]=pi/4`

Explanation

Let `sin ^{-1}(3 / 5)=theta`

Then `sin theta=frac{3}{5}`, where `theta in[frac{-pi}{2}, frac{pi}{2}]`

Thus, `tan theta=frac{3}{4}`, which gives `theta=tan ^{-1} frac{3}{4}`

Therefore. `2 sin ^{-1} frac{3}{5}-tan ^{-1} frac{17}{31}`


`=2 theta-tan ^{-1} frac{17}{31}=2 tan ^{-1} frac{3}{4}-tan ^{-1} frac{17}{31} `

`=tan ^{-1}(frac{2 cdot frac{3}{4}}{1-frac{9}{16}})-tan ^{-1} frac{17}{31}=tan ^{-1} frac{24}{7}-tan ^{-1} frac{17}{31} `

`=tan^{-1}(frac{frac{24}{7}-frac{17}{31}}{1+frac{24}{7}cdot frac{17}{31}})=frac{pi}{4}`