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

Simplity tan{1/2sin^(-1)((2x)/(1+x^(2)))+1/2cos^(-1)(1-y^(2))/(1+y^(2))} , if xgtygt1 .

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 .

Find the value of tan { 1/2 sin^(-1) ((2x)/(1+x^(2))) + 1/2 cos^(-1) ((1-y^(2))/(1+ y^(2)))}

If 1/2"sin"^(-1)(2x)/(1+x^2)+1/2"cos"^(-1)(1-y^2)/(1+y^2)+1/3"tan"^(-1)(3z-z^3)/(1-3z^2)=5pi then x+y+z=

If sin^(-1)((2x)/(1+x^(2)))+cos^(-1)((1-x^(2))/(1+x^(2)))=4 tan^(-1) x then

If y=sin^(-1)((1-x^(2))/(1+x^(2)))+cos^(-1)((1-x^(2))/(1+x^(2))) find (dy)/(dx)

sin{tan^(-1)[(1-x^(2))/(2x)]+cos^(-1)[(1-x^(2))/(1+x^(2))]}=

sin [ tan^(-1). (1 - x^(2))/(2x) + cos^(-1) . (1-x^(2))/(1 + x^(2))] is

Prove that : tan^(-1)((x-y)/(1+xy)) + tan^(-1)((y-z)/(1+yz)) + tan^(-1)( (z-x)/(1+zx)) = tan^(-1)((x^2-y^2)/(1+x^2y^2))+tan^(-1)((y^2-z^2)/(1+y^2z^2))+tan^(-1)((z^2-x^2)/(1+z^2x^2))