Prove that `tan{1/2sin^(-1)((2x)/(1+x^(2)))+1/2cos^(-1)((1-y^(2))/(1+y^(2)))}=(x+y)/(1-xy)`, where `|x| lt 1,y gt 0` and `xy lt 1`.

Explanation

Putting, `x=tantheta` and `y=tan phi`, we get
LHS `=tan{1/2sin^(-1)(2x)/(1-x^(2))+1/2cos^(-1)((1-y^(2))/(1+y^(2)))}`
`=tan{1/2sin^(-1)(2tantheta)/(1-tan^(2)theta)+1/2cos^(-1)((1-tan^(2)phi)/(1+tan^(2)phi))}`
`=tan{1/2sin^(-1)(sin2theta)+1/2cos^(-1)(cos2phi)}`
`=tan{(1/2 xx 2theta)+(1/2 xx 2phi)}`
`therefore tan{1/2sin^(-1)((2x)/(1-x^(2)))+1/2cos^(-1)((1-y^(2))/(1+y^(2))}=(x+y)/(1-xy)` = RHS