Suppose that, #arccos(x/a)=theta and arccos(y/b)=phi#.
#:. costheta=x/a and cosphi=y/b...........(ast)#.
#:. sintheta=sqrt(1-x^2/a^2) and sinphi=sqrt(1-y^2/b^2)...(ast^1)#.
We are given that, #theta+phi=alpha#.
#:. sinalpha=sin(theta+phi)#,
#=sinthetacosphi+costhetasinphi#,
#rArr sinalpha=sqrt(1-x^2/a^2)*y/b+x/a*sqrt(1-y^2/b^2)#.
#:. sin^2alpha=(1-x^2/a^2)y^2/b^2+x^2/a^2(1-y^2/b^2)#
#+(2xy)/(ab)*sqrt(1-x^2/a^2)*sqrt(1-y^2/b^2)#,
#={y^2/b^2-(x^2y^2)/(a^2b^2)}+{x^2/a^2-(x^2y^2)/(a^2b^2)}#
#+(2xy)/(ab)*sqrt(1-x^2/a^2)*sqrt(1-y^2/b^2)#,
#=x^2/a^2+y^2/b^2-(2x^2y^2)/(a^2b^2)+(2xy)/(ab)*sqrt(1-x^2/a^2)*sqrt(1-y^2/b^2)#,
#=x^2/a^2+y^2/b^2-(2xy)/(ab){(xy)/(ab)-sqrt(1-x^2/a^2)*sqrt(1-y^2/b^2)}#,
#=x^2/a^2+y^2/b^2-(2xy)/(ab){costhetacosphi-sinthetasinphi}...[because, (ast) and (ast^1)]#,
#=x^2/a^2+y^2/b^2-(2xy)/(ab)cos(theta+phi)#.
# rArr sin^2alpha=x^2/a^2+y^2/b^2-(2xy)/(ab)cos(alpha)#, as desired!
Enjoy Maths.!
