Answer
$ \dfrac{1}{\cos{\theta}\sin{\theta}}$
Work Step by Step
Combine the expressions by using their LCD, which is $\cos{\theta}\sin{\theta}$, to obtain:
\begin{align*}
\dfrac{\sin{\theta}+\cos{\theta}}{\cos{\theta}}+\dfrac{\cos{\theta}-\sin{\theta}}{\sin{\theta}} &= \dfrac{(\sin{\theta}+\cos{\theta})\sin{\theta}}{\cos{\theta}\sin{\theta}}+\dfrac{(\cos{\theta}-\sin{\theta})\cos{\theta}}{\cos{\theta} \sin{\theta}}\\\\
&= \dfrac{\sin^2{\theta}+\sin{\theta} \cos{\theta}}{\cos{\theta}\sin{\theta}}+\dfrac{\cos^{2}{\theta}-\sin{\theta}\cos{\theta}}{\cos{\theta}\sin{\theta}}\\\\
&= \dfrac{\sin^2{\theta}+\sin{\theta} \cos{\theta}+\cos^{2}{\theta}-\sin{\theta}\cos{\theta}}{\cos{\theta}\sin{\theta}}\\\\
&= \dfrac{\sin^2{\theta}+\cos^2{\theta}}{\cos{\theta}\sin{\theta}}
\end{align*}
Since $\sin^2{\theta}+\cos^2{\theta} = 1$, then
$\dfrac{\sin^2{\theta}+\cos^2{\theta}}{\cos{\theta}\sin{\theta}} =\boxed{ \dfrac{1}{\cos{\theta}\sin{\theta}}}$
