## Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

We know that $\sec \theta =\dfrac{1}{\cos \theta}$ and $\csc \theta=\dfrac{1}{\sin \theta}$ In order to prove the given identity, we simplify the left hand side $\text{LHS}$ as follows: $\text{LHS } =\dfrac{\sec \theta}{\csc \theta} +\dfrac{\sin \theta}{\cos \theta} \\= \dfrac{1/\cos \theta}{1/\sin \theta} +\dfrac{\sin \theta}{\cos \theta} \\ =\dfrac{\sin }{\cos \theta} +\dfrac{\sin \theta}{\cos \theta} \\=\tan \theta + \tan \theta \\= 2 \tan \theta \\= \text{ RHS}$