## Precalculus (6th Edition) Blitzer

The given expression on the left side $\frac{1}{\sin \theta +\cos \theta }+\frac{1}{\sin \theta -\cos \theta }$ can be simplified by rationalizing each term and then using the algebraic identity $\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$ for simplifying the terms. \begin{align} & \frac{1}{\sin \theta +\cos \theta }+\frac{1}{\sin \theta -\cos \theta }=\frac{\sin \theta -\cos \theta }{\sin \theta -\cos \theta }.\frac{1}{\sin \theta +\cos \theta } \\ & +\frac{\sin \theta +\cos \theta }{\sin \theta +\cos \theta }.\frac{1}{\sin \theta -\cos \theta } \\ & =\frac{\sin \theta -\cos \theta }{{{\sin }^{2}}\theta -{{\cos }^{2}}\theta }+\frac{\sin \theta +\cos \theta }{{{\sin }^{2}}\theta -{{\cos }^{2}}\theta } \\ & =\frac{\sin \theta -\cos \theta +\sin \theta +\cos \theta }{{{\sin }^{2}}\theta -{{\cos }^{2}}\theta } \\ & =\frac{2\sin \theta }{{{\sin }^{2}}\theta -{{\cos }^{2}}\theta } \end{align} And the expression can be further simplified by rationalizing it again: \begin{align} & \frac{2\sin \theta }{{{\sin }^{2}}\theta -{{\cos }^{2}}\theta }=\frac{2\sin \theta }{{{\sin }^{2}}\theta -{{\cos }^{2}}\theta }.\frac{{{\sin }^{2}}\theta +{{\cos }^{2}}\theta }{{{\sin }^{2}}\theta +{{\cos }^{2}}\theta } \\ & =\frac{2\sin \theta .1}{{{\sin }^{4}}\theta -{{\cos }^{4}}\theta } \\ & =\frac{2\sin \theta }{{{\sin }^{4}}\theta -{{\cos }^{4}}\theta } \end{align} Hence, the expression on the left side is equal to the expression on the right side.