सिद्ध कीजिए -
`tan^(-1)((sqrt(1+x)-sqrt(1-x))/(sqrt(1+x)+sqrt(1-x)))=(pi)/(4)-(1)/(2)cos^(-1)x, -(1)/(sqrt2)le x le1.`

Explanation

`L.H.S.=tan^(-1)((sqrt(1+x)-sqrt(1-x))/(sqrt(1+x)+sqrt(1-x)))`
`x=cos2theta` रखने पर,
`=tan^(-1)[(sqrt(1+cos2theta)-sqrt(1-cos 2theta))/(sqrt(1+cos 2theta)+sqrt(1-cos 2theta))]`
`=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)))],`
`" "[because cos 2theta=2cos^(2)theta-1=1-2sin^(2)theta]`
`=tan^(-1)[(sqrt(2cos^(2)theta)-sqrt(2sin^(2)theta))/(sqrt(2cos^(2)theta)+sqrt(2sin^(2)theta))]`
`=tan^(-1)[(sqrt2cos theta-sqrt2sin theta)/(sqrt2cos theta+sqrt2sin theta)]`
`=tan^(-1)[(cos theta=sin theta)/(cos theta+sin theta)]`
`=tan^(-1)[(1-tan theta)/(1+tan theta)],`
[अंश और हर को `cos theta` से भाग देने पर ]
`=tan^(-1)[(tan.(pi)/(4)-tantheta)/(1+tan.(pi)/(4).tan theta)],`
`" "[because tan (A-B)=(tanA-tanB)/(1+tan Atan B)]`
`=tan^(-1)[tan((pi)/(4)-theta)]`
`=(pi)/(4)-theta`
`=(pi)/(4)-(1)/(2)cos^(-1)x`
`[because x=cos 2theta rArr 2 theta cos^(-1)x rArr theta=(1)/(2)cos^(-1)x]" "=(x)/(2)`
= R.H.S. यही सिद्ध करना था ।