if tan inverse[{(1+x²)^1/2 - (1-x²)^1/2} / {((1+x²)^1/2 + (1-x²)^1/2)} = a then x²=________

ans: sin2a

let tan^-1{(1+x²)^1/2- (1-x²)^1/2}/(1+x²)^1/2+(1-x²)^1/2  = a

then

[{(1+x²)^1/2 - (1-x²)^1/2} / {((1+x²)^1/2 + (1-x²)^1/2) = tan a

Now apply C and D to get,

{1-x^2/1+x²}^1/2  =  1-tana/1+tana

expand RHS by writing tan a = sina/cosa.

squaring,

{1-x^2/1+x²}  = {cosa - sina/cosa + sina}²

{1-x^2/1+x²} = 1- sin2a/1+sin2a

so x² =sin2a.