Answer 1

Let x = tan θ. Then, θ = tan⁻¹ x.

`:. sin^(-1)  (2x)/(1+x^2 ) = sin^(-1)  ((2tan theta)/(1 + tan^2 theta)) = sin^(-1) (sin 2 theta) = 2theta = 2 tan^(-1) x`

Let y = tan Φ. Then, Φ = tan⁻¹ y.

`:. cos^(-1)  (1 - y^2)/(1+ y^2) = cos^(-1) ((1 - tan^2 phi)/(1+tan^2 phi)) =    cos^(-1)(cos 2phi) = 2phi = 2 tan^(-1) y`

`:. tan  1/2 [sin^(-1)  "2x"/(1+x^2) + cos^(-1)  (1-y^2)/(1+y^2)]`

`= tan  1/2 [2tan^(-1) x + 2tan^(-1) y]`

`= tan[tan^(-1) x + tan^(-1) y]`

`= tan[tan^(-1) ((x+y)/(1-xy))]`

`= (x+y)/(1-xy)`