## Precalculus (6th Edition) Blitzer

Let us consider the left side of the given expression: $\tan \left( \frac{\pi }{4}-\theta \right)$ By using the identities of trigonometry, $\sin \,\left( \alpha -\beta \right)=\sin \,\alpha \,\cos \beta -\cos \,\alpha \,\sin \,\beta$ $\cos \,\left( \alpha -\beta \right)=\cos \,\alpha \,\cos \beta +\sin \,\alpha \,\sin \,\beta$ and $\tan \,\alpha =\frac{\sin \,\alpha }{\cos \,\alpha }$ , the above expression can be further simplified as: \begin{align} & \tan \left( \frac{\pi }{4}-\theta \right)=\frac{\sin \left( \frac{\pi }{4}-\theta \right)}{\cos \left( \frac{\pi }{4}-\theta \right)} \\ & =\frac{\sin \,\frac{\pi }{4}\,\cos \,\theta -\cos \,\frac{\pi }{4}\,\sin \,\theta }{\cos \,\frac{\pi }{4}\,\cos \,\theta +\sin \,\frac{\pi }{4}\,\sin \,\theta } \\ & =\frac{\cos \,\theta \times \frac{1}{\sqrt{2}}-\sin \,\theta \times \frac{1}{\sqrt{2}}}{\cos \,\theta \times \frac{1}{\sqrt{2}}+\sin \,\theta \times \frac{1}{\sqrt{2}}} \\ & =\frac{\frac{1}{\sqrt{2}}\times \left( \cos \,\theta -\sin \,\theta \right)}{\frac{1}{\sqrt{2}}\times \left( \cos \,\theta +\sin \,\theta \right)} \end{align} $=\frac{\cos \,\theta -\sin \,\theta }{\cos \,\theta +\sin \,\theta }$ Hence, the left side of the given expression is equal to the right side, which is $\tan \left( \frac{\pi }{4}-\theta \right)=\frac{\cos \,\theta -\sin \,\theta }{\cos \,\theta +\sin \,\theta }$.