If `tan^(-1){(sqrt(1+x^2)-sqrt(1-x^2))/(sqrt(1+x^2)+sqrt(1-x^2))}=alpha,` then prove that `x^2=sin2alpha`

Explanation

Let, `tan^(-1)[(sqrt(1+x^(2))+sqrt(1-x^(2)))/(sqrt(1+x^(2))-sqrt(1-x^(2)))], -1ltxlt1, x ne 0`
Put `x^(2)=cos2theta`
`:. y = tan^(-1)'((sqrt(1+cos2theta)+sqrt(1-cos2theta))/(sqrt(1+cos2theta)-sqrt(1-cos2theta)))`
`=tan^(-1)((sqrt(1+2cos^(2)theta-1)+sqrt(1-1+2sin^(2)theta))/(sqrt(1+2cos^(2)theta-1)-sqrt(1-1+2sin^(2)theta)))`
`= tan^(-1)'((sqrt(2)costheta+sqrt(2)sintheta)/(sqrt(2)costheta-sqrt(2)sin theta))= tan^(-1)[(sqrt(2)
(costheta+sintheta))/(sqrt(2)(costheta-sintheta))]`
` =tan^(-1)((costheta+sintheta)/(costheta-sintheta))=tan^(-1)'(((costheta+sintheta)/(costheta))/((costheta-sintheta)/(costheta)))`
`= tan^(-1)'((1+tantheta)/(1-tantheta))`
`= tan^(-1)((1+tantheta)/(1-tantheta))`
`= tan^(-1)tan'((pi)/(4)+theta), [:'tan(a+b)=(tana+tanb)/(1-tana.tanb)]`
`= (pi)/(4)+theta = (pi)/(4)+(1)/(2)cos^(-1)x^(2) , [:' 2theta=cos^(-1)x^(2) rArr theta = (1)/(2) cos^(-1) x^(2)]`
`:. (dy)/(dx) = (d)/(dx)(pi/4)+(d)/(dx)(1/2 cos^(-1)x^(2))`
`= 0+(1)/(2).(-1)/(sqrt(1-x^(4))).(d)/(dx)x^(2)=(1)/(2).(-2x)/(sqrt(1-x^(4)))= (-x)/(sqrt(1-x^(4)))`.