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

$\dfrac{1+\sin{\theta}}{\cos{\theta}}$
Recall: $(a-b)(a+b)=a^2-b^2$ Use the rule above to obtain: \begin{align*} \dfrac{\cos{\theta}}{1-\sin{\theta}} \times \dfrac{1+\sin{\theta}}{1+\sin{\theta}}&= \dfrac{\cos{\theta}(1+\sin{\theta})}{(1-\sin{\theta})(1+\sin{\theta})}\\\\ &= \dfrac{\cos{\theta}(1+\sin{\theta})}{1-\sin^2{\theta}} \end{align*} Since $1-\sin^2{\theta} = \cos^2{\theta}$, then: $\dfrac{\cos{\theta}(1+\sin{\theta})}{1-\sin^2{\theta}} = \dfrac{\cos{\theta}(1+\sin{\theta})}{\cos^2{\theta}} = \boxed{\dfrac{1+\sin{\theta}}{\cos{\theta}}}$