# Solution

Determine if the following is an identity: $$\frac{1+\tan t}{\tan t} = \cot t + \sec^2 t - \tan^2 t$$

Let us start by trying to simplify the left side, $$\begin{array}{rclc} \frac{1+\tan t}{\tan t} &=& \frac{\displaystyle{1+\frac{\sin t}{\cos t}}}{\displaystyle{\frac{\sin t}{\cos t}}} & \quad \scriptsize{\textrm{rewriting in terms of } \sin \theta \textrm{ and } \cos \theta}\\\\ &=& \frac{\displaystyle{1+\frac{\sin t}{\cos t}}}{\displaystyle{\frac{\sin t}{\cos t}}} \cdot \frac{\cos t}{\cos t} & \overset{\normalsize{\textrm{to simplify the complex fraction}}}{\scriptsize{\textrm{we multiply by a well-chosen value of one}}}\\\\ &=& \frac{\cos t + \sin t}{\sin t} \end{array}$$

With it unclear how to further simplify the left side, we turn to simplifying the right side -- attempting to match the last expression above:

$$\begin{array}{rclc} \cot t + \sec^2 t - \tan^2 t &=& \frac{\cos t}{\sin t} + \frac{1}{\cos^2 t} - \frac{\sin^2 t}{\cos^2 t} & \quad \scriptsize{\textrm{rewriting in terms of } \sin \theta \textrm{ and } \cos \theta}\\\\ &=& \frac{\cos t}{\sin t} + \frac{1 - \sin^2}{\cos^2 t} & \overset{\normalsize{\textrm{note the left-most term matches part of } \frac{\cos t + \sin t}{\sin t}}}{\scriptsize{\textrm{so we only combine the other two terms...}}}\\\\ &=& \frac{\cos t}{\sin t} + \frac{\cos^2 t}{\cos^2 t} & \scriptsize{\textrm{recalling} \sin^2 \theta + \cos^2 \theta = 1}\\\\ &=& \frac{\cos t}{\sin t} +1 & \overset{\normalsize{\textrm{now we combine fractions as the side we are}}}{\scriptsize{\textrm{attempting to match is a fraction, not a sum}}}\\\\ &=& \frac{\cos t}{\sin t} +\frac{\sin t}{\sin t}\\\\ &=& \frac{\cos t + \sin t}{\sin t} \end{array}$$

Since this just happens to match the simplified version of the left side, we now know that the equation above is an identity.