# (1-2sinx-3sin^2x)/cos^2x=(1-3sinx)/(1-sinx) prove the identity

When verifying an identity, you cannot use the assumption that the two sides are equal. You must work on one side and show, via correct algebraic manipulations, that it is equivalent to the other side.

Verify `(1-2sinx-3sin^2x)/(cos^2x)=(1-3sinx)/(1-sinx)` :

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

`=(-(-1+2sinx+3sin^2x))/(1-sin^2x)` Factor out a -1;Use Pythagorean identity in the denominator

`=(-(-1+3sinx)(1+sinx))/((1+sinx)(1-sinx))` Factor the numerator in terms of sinx

`=(1-3sinx)/(1-sinx)` Cancel

`=RHS` as required.

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You need to multiply both sides by `cos^2 x`  such that:

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

You need to multiply both sides by `(1-sinx)`  such that:

`(1-sinx)(1-2sinx-3sin^2x) = cos^2 x(1-3sinx) `

You need to use the fundamental formula of trigonometry such that:

`1 - sin^2 x = cos^2 x `

Substituting `1 - sin^2 x`  for `cos^2 x`  yields:

`(1-sinx)(1-2sinx-3sin^2x) = (1 - sin^2 x)(1-3sinx) `

Opening the brackets yields:

`1 - 2sin x - 3sin^3 x - sin x + 2sin^2 x + 3sin^3 x = 1 - 3sin x - sin^2 x + 3sin^3 x`

Reducing like terms yields:

`-3 sin x+ 2sin^2 x = - 3sin x - sin^2 x + 3sin^3 x`

`3sin^2 x = 3sin^3 x => sin^2 x = sin^3 x`

Notice that the identity is valid only for the following values of x such that `x in {0, pi/2, pi, 2pi,}.`