## Precalculus (6th Edition) Blitzer

Considered the identity, $\tan \theta +\cot \theta =\sec \theta \csc \theta$ Now, apply the Quotient identity $\tan \theta =\frac{\sin \theta }{\cos \theta }$ and $\cot \theta =\frac{\cos \theta }{\sin \theta }$ , \begin{align} & \tan \theta +\cot \theta =\frac{\sin \theta }{\cos \theta }+\frac{\cos \theta }{\sin \theta } \\ & =\frac{\sin \theta \cdot \sin \theta +\cos \theta \cdot \cos \theta }{\sin \theta \cos \theta } \\ & =\frac{{{\sin }^{2}}\theta +{{\cos }^{2}}\theta }{\sin \theta \cos \theta } \end{align} Then, use the Pythagorean identity ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$ , $\tan \theta +\cot \theta =\frac{1}{\sin \theta \cos \theta }$ Now, use the Reciprocal identity $\csc \theta =\frac{1}{\sin \theta }$ and $\sec \theta =\frac{1}{\cos \theta }$ , \begin{align} & \tan \theta +\cot \theta =\left( \frac{1}{\cos \theta } \right)\left( \frac{1}{\sin \theta } \right) \\ & =\sec \theta \csc \theta \end{align} Hence, the left side is identical to the right side $\tan \theta +\cot \theta =\sec \theta \csc \theta$.