# How do I prove that (3-6 cos^2 x)/(sin x- cos x) = 3(sin x + cos x)?

##### 1 Answer
Sep 20, 2015

Yes, the equality is correct. See explanation.

#### Explanation:

You have
$\frac{3 - 6 {\cos}^{2} x}{\sin x - \cos x}$

On the numerator, factor out 3:
$= 3 \cdot \frac{1 - 2 {\cos}^{2} x}{\sin x - \cos x}$

Remember the identity ${\sin}^{2} x + {\cos}^{2} x = 1$.
Substitute the 1 in the numerator:
$= 3 \cdot \frac{{\sin}^{2} x + {\cos}^{2} x - 2 {\cos}^{2} x}{\sin x - \cos x}$

Simplify by adding the cos^2 x together:
$= 3 \cdot \frac{{\sin}^{2} x - {\cos}^{2} x}{\sin x - \cos x}$

Remember that when you have something of the form ${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$

So rewrite the numerator
$= 3 \cdot \frac{\left(\sin x - \cos x\right) \left(\sin x + \cos x\right)}{\sin x - \cos x}$

And remove $\sin x - \cos x$ from the numerator and denominator. You are left with:

$= 3 \left(\sin x + \cos x\right)$

Q.E.D.