Simplify $\frac{\sin t - \cos t}{\sin t + \cos t}$ $$\frac{\sin t - \cos t}{\sin t + \cos t}$$
I am to simplify this into $$\tan(t - \frac{\pi}{4})$$
I'm not sure how to carry on, though. Multiplying numerator and denominator by denominator (or numerator) doesn't get me anywhere.
 A: You know that you already need to use an angle addition identity, because the desired result $\tan (t - \pi/4)$ contains the difference of two angles.
So, which such identities do you know?  Specifically, do you know of the tangent angle addition identity $$\tan (a - b) = \frac{\sin (a - b)}{\cos (a - b)} = \frac{\sin a \cos b - \cos a \sin b}{\cos a \cos b + \sin a \sin b}?$$  Now, what happens if we let $a = t$ and $b = \pi/4$?
A: Step-by-step
$$ \dfrac{\sin t - \cos t}{\sin t + \cos t}$$
$$= \dfrac{\dfrac{\sin t}{\cos t} - \dfrac{\cos t}{\cos t}}{\dfrac{\sin t}{\cos t} + \dfrac{\cos t}{\cos t}}$$
$$= \dfrac{\tan t - 1}{\tan t + 1}$$
$$= \dfrac{\tan t - 1}{1 + \tan t \color{blue}{\cdot 1}}$$
Substitute $\color{blue}{\tan \frac{\pi}{4} = 1}$
$$= \dfrac{\tan t - \tan \frac{\pi}{4}}{1 + \tan t \color{blue}{\tan \frac{\pi}{4}}}$$
$$= \tan (t - \frac{\pi}{4})$$
We have used the formula
$$ \tan(A-B) = \dfrac{\tan A - \tan B}{1 + \tan A \tan B} $$
A: Note that $$\frac{\sin t - \cos t}{\sin t + \cos t} = \frac{\sqrt{2}\sin(t - \frac{\pi}{4})}{\sqrt{2}\sin(t + \frac{\pi}{4})} =  \frac{\sqrt{2}\sin(t - \frac{\pi}{4})}{\sqrt{2}\cos(t - \frac{\pi}{4})} = \tan(t- \frac{\pi}{4})$$
