`tan^(-1)((sqrt(1+x)-sqrt(1-x))/(sqrt(1+x)+sqrt(1-x)))=(pi)/(4)-(1)/(2)cos^(-1)x,-(1)/(sqrt2)lexle1`

Explanation

माना `x=costhetaimpliestheta=cos^(-1)x`
`L.H.S.=tan^(-1)((sqrt(1+x)-sqrt(1-x))/(sqrt(1+x)+sqrt(1-x)))`
`=tan^(-1)((sqrt(1+costheta)-sqrt(1-costheta))/(sqrt1+costheta+sqrt(1-costheta)))`
`=tan^(-1)""(sqrt(2cos^(2)""(theta)/2)-sqrt(2sin^(2)""(theta)/(2)))/(sqrt(2cos^(2)""(theta)/(2))+sqrt(2sin^(2)""(theta)/(2)))`
`=tan^(-1)((cos""(theta)/(2)-sin""(theta)/(2))/(cos""(theta)/(2)+sin""(theta)/(2)))`
`=tan^(-1)((1-tan""(theta)/(2))/(1+tan""(theta)/(2)))`
`=tan^(-1)tan((pi)/(4)-(theta)/(2))=(pi)/(4)-(theta)/(2)`
`=(pi)/(4)-(1)/(2)cos^(-1)x=R.H.S.`