Solution

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


Let us start by trying to simplify the left side,$\require{cancel}$

$$\begin{array}{rclc} \frac{\tan^2 t - 1}{\sin t - \cos t} &=& \frac{\displaystyle{\frac{\sin^2 t}{\cos^2 t} - 1}}{\sin t - \cos t} & \quad \scriptsize{\textrm{rewriting in terms of } \sin \theta \textrm{ and } \cos \theta}\\\\ &=& \frac{\displaystyle{\frac{\sin^2 t}{\cos^2 t} - 1}}{\sin t - \cos t} \cdot \frac{\cos^2 t}{\cos^2 t} & \overset{\normalsize{\textrm{to simplify the complex fraction}}}{\scriptsize{\textrm{we multiply by a well-chosen value of one}}}\\\\ &=& \frac{\sin^2 t - \cos^2 t}{\cos^2 (\sin t - \cos t)} & \overset{\normalsize{\textrm{we don't distribute the factor of } \cos^2 t}}{\scriptsize{\textrm{as it appears on the right side of the equation we are checking}}}\\\\ &=& \frac{(\sin t + \cos t)(\sin t - \cos t)}{\cos^2 t (\sin t - \cos t)} & \overset{\normalsize{\textrm{factoring things in the hope that we can}}}{\scriptsize{\textrm{eliminate the extra factor in the denominator }}}\\\\ &=& \frac{\sin t + \cos t}{\cos^2 t} \cdot \cancel{\frac{\sin t - \cos t}{\sin t - \cos t}} & \quad \scriptsize{\textrm{...and it does!}}\\\\ \end{array}$$

At this point, the similarity with the right-hand side of the given equation causes us to pause...

Note that if $\cos t \neq 0$, this expression will have a different value than the right-hand side of the given equation. To demonstrate a specific counter-example, consider $t = 0$:

$$\frac{\tan^2 0 - 1}{\sin 0 - \cos 0} = 0, \quad \textrm{ while } \quad \frac{\sin 0 - \cos 0}{\cos^2 0} = -1$$

Thus, the given equation is not and identity