The general solution of the equation `(1-sinx+....+(-1)^n sin^n x+...)/(1+sinx+......+sin^nx+...)= (1-cos2x)/(1+cos2x)` is

Options

A. `(-1)^(n)(pi//6)+n pi`
B. `(-1)^(n)(pi//3)+n pi`
C. `(-1)^(n+1)(pi//6)+ n pi`
D. `(-1)^(n-1)(pi//3)+n pi,(n in 1)`

Answer

A

Explanation

The equation
`(1-sin x +…..+(-1)^(n)sin^(n)x +….)/(1+sin x + ….+ sin^(n) x +….)=(1-cos 2x)/(1+ cos 2x)`
`rArr (1)/(1+sin x)xx(1-sin x)/(1)=(2 sin^(2)x)/(2 cos^(2)x)`
`rArr (1-sin x)/(1+sin x)=(sin^(2)x)/(1-sin^(2)x)`
`rArr (1-sin x)^(2)=sin^(2)x`
`rArr sin x=(1)/(2)`
`rArr sin x = sin (pi//6)`
`rArr x = n pi+(-1)^(n)(pi//6), n in Z`