`tan^(-1) [(cos x)/(1-sinx)]`

Let f(x) = tan^(-1) ((1+cos x)/(sinx)) and g (x) = tan^(-1) ((sinx)/(1-cos x)) int {f(x) - g (x)}dx =

Let f(x) = tan^(-1) ((1+cos x)/(sinx)) and g (x) = tan^(-1) ((sinx)/(1-cos x)) int {f(x)+g(x)}dx =

Let f(x) = tan^(-1) ((1+cos x)/(sinx)) and g (x) = tan^(-1) ((sinx)/(1-cos x)) int f(x) dx =

Find the derivative with tan^(-1) ((sinx )/(1 + cos x)) with respect to tan^(-1) ((cos x)/(1 + sin x))

Express each of the following in the simplest form: tan^(-1){(cosx)/(1-sinx)},\ -pi/2

Let y=tan ^(-1)((1-cos x)/(sin x)) i) Express 1-cos x and sinx in terms of sin (x/2) and cos (x/2) ii) Hence find (dy/dx) .

Solve tan^(-1)""[(a cos x -b sinx)/(b cosx+a sinx)] , if ""(a)/(b)tanx le-1

Differentiate each of the following w.r.t. x: (i) tan^(-1)((1-cosx)/(sinx))" "(ii)tan^(-1)((cos x -sinx)/(cosx +sinx))

If f (x) = tan ^(-1) ((cos x- sinx)/(cos x+ sin x)) then the value of (d)/(dx) f(x) is-

tan ^ (- 1) ((cos x-sin x) / (cos x + sin x)) = tan ^ (- 1) ((1-tan x) / (1 + tan x))