Parametric differentiation

Question

The parametric equations of a curve are $$\begin{cases}x(t)=e^{-t}\cos t\\y(t)=e^{-t}\sin t\end{cases}$$

Show that $$\frac{dy}{dx}= \tan\left(t-\frac{\pi}{4}\right)$$

I did the differentiation correct which is

$$\frac{\sin t-\cos t}{\cos t+\sin t}$$

but I don't know how can I reach the final answer? how can this be changed to $\tan\left(t-\dfrac{\pi}{4}\right)$

Answer 1

$$\frac{\sin t-\cos t}{\cos t+\sin t}=\frac{\sqrt 2(\sin t\,\cos \frac\pi 4-\sin\frac\pi 4\,\cos t)}{\sqrt 2(\cos t\,\cos \frac\pi 4+\sin t\,\sin\frac\pi 4)}=\frac{\sin(t-\frac\pi 4)}{\cos(t-\frac\pi 4)}.$$

Answer 2

Use the formula $\tan(a-b)$:

$$\tan(a-b) = \frac{\tan(a) -\tan(b)}{1 + \tan(a)\tan(b)}$$

Substituting this in, we get:

$$\tan(t-\pi/4) = \frac{\tan(t) -\tan\left(\frac{\pi}{4}\right)}{1 + \tan(t)\tan\left(\frac{\pi}{4}\right)} = \frac{\tan(t) - 1}{1 + \tan(t)} =\frac{\cfrac{\sin(t) -\cos(t)}{\cos(t)}}{\cfrac{\sin(t) +\cos(t)}{\cos(t)}} = \frac{\sin(t) - \cos(t)}{\sin(t) + \cos(t)}$$

This equals that initial answer you arrived at when differentiating the parametric equation.