## Precalculus (6th Edition) Blitzer

Consider the given expression $\frac{\cos 2x}{\cos x-\sin x}$ of the identity and solve it. Recall the identity $\cos 2x={{\cos }^{2}}x-{{\sin }^{2}}x$ and the expansion of ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$. \begin{align} & \frac{\cos 2x}{\cos x-\sin x}=\frac{{{\cos }^{2}}x-{{\sin }^{2}}x}{\cos x-\sin x} \\ & =\frac{\left( \cos x+\sin x \right)\left( \cos x-\sin x \right)}{\cos x-\sin x} \\ & =\cos x+\sin x \end{align} Hence, proved.