# Prove `tan2x cosec x = sin x / (1-sin 2x).`

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You should remember that `csc x = 1/sin x` , hence, substituting `1/sin x` for `csc x ` yields:

`(tan 2x)/(sin x) = (sin x)/(1 - sin 2x) => tan 2x*(1 - sin 2x) = sin^2 x`

`tan 2x - tan 2x*sin 2x = sin^2 x`

`tan 2x = sin^2 x + (sin^2(2x))/(cos 2x)`

`tan 2x = sin^2 x + 1/(cos 2x) - cos 2x`

`tan 2x = sin^2 x + 1/(cos 2x) - 1 + 2sin^2 x`

`tan 2x = 3sin^2 x + 1/(cos 2x) - 1`

`tan 2x = (3sin^2 x(1 - 2sin^2 x) + 1 - 1 + 2sin^2 x)/(cos 2x)`

`tan 2x = (5sin^2 x - 6sin^4 x)/(cos 2x)`

Substituting `(sin 2x)/(cos 2x) ` for `tan 2x` yields:

`(sin 2x)/(cos 2x) = (5sin^2 x - 6sin^4 x)/(cos 2x)`

`sin 2x = 5sin^2 x - 6sin^4 x` invalid

**Notice that using the information provided by the problem yields an invalid result, hence `tan 2x csc x != sin x/(1 - sin 2x).` **