Differentiate `tan^(-1){sqrt((1-cosx)/(1+cosx))},\ -pi

Answer

N/a

Explanation

Let `y = tan^(-1) sqrt((1-cosx)/(1+cosx))`
`:. (dy)/(dx)=d/dxtan^(-1)sqrt((1-cosx)/(1+cosx))`
`= (1)/(1+sqrt(((1-cosx)/(1+cosx))^(2))).d/dx[(1-cosx)/(1+cosx)]^(1//2), [:'(d)/(dx)(tan^(-1)x) = (1)/(1+x^(2))]`
`= (1)/(1+(1-cosx)/(1+cosx)).1/2[(1-cosx)/(1+cosx)]^(1//2) . (d)/(dx) ((1-cosx)/(1+cox))`
`= (1)/((1+cosx+1-cosx)/(1+cosx)).1/2[((1-cosx))/((1+cosx)).((1-cosx))/((1-cosx))]^(-1//2).((1+cosx).sinx+(1-cosx).sinx)/((1+cosx^(2)))`
`= ((1+ cosx))/(2).(1)/(2)[((1-cosx)^(2))/((1-cos^(2)x))]^(1//2)[(sinx(1+cosx+1-cosx))/((1+cosx)^(2))]`
`= ((1+ cosx))/(2).(1)/(2)[((1-cosx)^(2))/((1-cos^(2)x))]^(1//2)[(sinx(1+cosx+1-cosx))/((1+cosx)^(2))]`
`= ((1+ cosx))/(2).(1)/(2)[((1-cosx)^(2))/((1-cos^(2)x))]^(1//2)[(sinx(1+cosx+1-cosx))/((1+cosx)^(2))]`
`= ((1+ cosx))/(2).(1)/(2)[((1-cosx)^(2))/((1-cos^(2)x))]^(1//2)[(sinx(1+cosx+1-cosx))/((1+cosx)^(2))]`
`= ((1+cosx))/(2).(1)/(2)[((1-cosx)^(2))/(sinx)]^(-1//12) .(2sinx)/((1+cosx)^(2))`
`= ((1+cosx))/(2).(1)/(2).(sinx)/((1-cosx)).(2sinx)/((1+cosx)^(2))`
`= (2sin^(2)x)/(4(1+cosx)(1-cosx))= (1)/(2).(sin^(2)x)/((1-cosx))`
`= 1/2.(sin^(2)x)/(sin^(2)x) = 1/2`
Alaternate Method
Let `y = tan^(-1)(sqrt((1-cosx)/(1+cosx)))`
`=tan^(-1)(sqrt((1-1+2sin^(2)'x/2)/(1+2cos^(2)'x/2-1))), [:' cosx = 1-2 sin^(2)'x/2=2cos^(2)'x/2-1]`
`=tan^(-1)(tan'x/2)=x/2`
On differentiating w.r.t. we get
`(dy)/(dx) = 1/2`