# How do you prove sinx+cosx=(2sin^2x-1)/(sinx-cosx)?

Apr 30, 2015

Try this:
$\sin x + \cos x = \frac{2 \left(1 - {\cos}^{2} x\right) - {\sin}^{2} x - {\cos}^{2} x}{\sin x - \cos x}$
$\left(\sin x + \cos x\right) \left(\sin x - \cos x\right) = 2 - 2 {\cos}^{2} x - {\sin}^{2} x - {\cos}^{2} x$
${\sin}^{2} x - \cancel{{\cos}^{2} x} = 2 - 2 {\cos}^{2} x - {\sin}^{2} x - \cancel{{\cos}^{2} x}$
$2 {\sin}^{2} x = 2 \left(1 - {\cos}^{2} x\right)$
$2 {\sin}^{2} x = 2 {\sin}^{2} x$

Where I used the fact that:
${\sin}^{2} x + {\cos}^{2} x = 1$