## Precalculus (6th Edition) Blitzer

$\frac{{{\sin }^{2}}x-{{\cos }^{2}}x}{\sin x+\cos x}$ By using the formulae $(A+B)(A-B)={{A}^{2}}-{{B}^{2}}$ , with $A=\sin x$ and $B=\cos x$ for the first numeric expression, we get: \begin{align} & \frac{{{\sin }^{2}}x-{{\cos }^{2}}x}{\sin x+\cos x}=\frac{\left( \sin x+\cos x \right)\left( \sin x-\cos x \right)}{\left( \sin x+\cos x \right)} \\ & =\sin x-\cos x \end{align} Hence, the the left side of the expression is equal to the right side, which is $\frac{{{\sin }^{2}}x-{{\cos }^{2}}x}{\sin x+\cos x}=\sin x-\cos x$.