Differentiate `tan^(-1){(sqrt(1+sinx)+sqrt(1-sinx))/(sqrt(1+sinx)-sqrt(1-sinx))}`, `0 < x < pi`

Explanation

We have
`y=tan^(-1){(sqrt(1+sinx)+sqrt(1-sinx))/(sqrt(1+sinx)-sqrt(1-sinx))}`, `0 < x < pi`
`y=tan^(-1){(sqrt(1+sinx)+sqrt(1-sinx))/(sqrt(1+sinx)-sqrt(1-sinx))xx(sqrt(1+sinx)+sqrt(1-sinx))/(sqrt(1+sinx)+sqrt(1-sinx))}`
`=>y=tan^(-1){((1+sinx)+(1-sinx)+2sqrt(1-sin^(2)x))/((1+sinx)-(1-sinx))}`
`=>y=tan^(-1)((1+cosx)/(sinx))`
`=>y=tan^(-1){(2cos^(2)(x//2))/(2sin(x//2)cos(x//2))}`
`=>y=tan^(-1){"cot"(x)/(2)}`
`=>y=tan^(-1){tan((pi)/(2)-(x)/(2))}`
`=((pi)/(2)-(x)/(2)).`
`therefore(dy)/(dx)=(d)/(dx)((pi)/(2)-(x)/(2))`
`=(d)/(dx)((pi)/(2))-(d)/(dx)((x)/(2))`
`=(0-(1)/(2))`
`=-(1)/(2).`