Respuesta :

We have the expression:

[tex]\frac{1-2\cos 2x}{2\sin x-1}[/tex]

Let's work first with the numerator. We have that we can write cos2x like this:

[tex]\cos 2x=1-\sin ^2x[/tex]

doing this substitution we get the following:

[tex]\begin{gathered} 1-2\cos 2x=1-2\cdot(1-\sin ^2x)=1-2+4\sin ^2x=4\sin ^2x-1 \\ =(2\sin x+1)(2\sin x-1) \end{gathered}[/tex]

Now that we have these two factors, we can use them on the original identity to get:

[tex]\frac{1-2\cos2x}{2\sin x-1}=\frac{(2\sin x+1)(2\sin x-1)}{2\sin x-1}=2\sin x+1[/tex]

therefore, the resulting identity is 2sinx+1

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