`tan^(-1)""(cosx)/(1-sinx),-(-3pi)/(2) lt x lt (pi)/(2)` , को सरलता रूप में व्यक्त कीजिए ।

Explanation

We have,
LHS `=tan^(-1)(cosx)/(1-sinx)`
`=tan^(-1){(sin(pi/2-x))/(1-cos(pi/2-x))}`
`=tan^(-1){(2sin(pi/4-x/2)cos(pi/4-x/2))/(2sin^(2)(pi/4-x/2)}`
`tan^(-1){cot(pi/4-x/2)}=tan^(-1)[tan{pi/4-x/2)}]`
`=tan^(-1){tan(pi/4+x/2)}`
`=(pi/4+x/2)`=RHS.
Hence, `tan^(-1)(cosx)/(1-sinx)=(pi/4+2/x)`.