Determine if the following is an identity: $$\displaystyle{\sin t \cos t = \frac{1}{\tan t + \cos t}}$$

There is not much we can do to simplify the left side, so we work with the right side first:

$$\begin{array}{rclc} \frac{1}{\tan t + \cos t} &=& \frac{1}{\displaystyle{\frac{\sin t}{\cos t} + \frac{\cos t}{\sin t}}} & \quad \scriptsize{\textrm{rewriting in terms of } \sin \theta \textrm{ and } \cos \theta}\\\\ &=& \frac{1}{\displaystyle{\frac{\sin^2 t}{\sin t \cos t} + \frac{\cos^2 t}{\sin t \cos t}}} & \overset{\normalsize{\textrm{getting common denominators}}}{\scriptsize{\textrm{so these fractions can be combined}}}\\\\ &=& \frac{1}{\displaystyle{\frac{\sin^2 t + \cos^2 t}{\sin t \cos t}}}\\\\ &=& \frac{1}{\displaystyle{\, \frac{1}{\sin t \cos t} \,}} & \scriptsize{\textrm{recalling} \sin^2 \theta + \cos^2 \theta = 1}\\\\ &=& \sin t \cos t & \scriptsize{\textrm{after simplifying by multiplying by a reciprical}} \end{array}$$

This last expression just happens to be identical to the left side of the equation we were given -- thus, the equation above is an identity.