How to convert this particular expression into some desired form?

Question

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

Show that $dy/dx =tan(t-\pi/4) $.

how to solve this? I can get a $dy/dx$ but i cannot convert into the desired form above.

This is the $dy/dx$ expression i get:

$dy/dx=\dfrac{Sin(t)-Cos(t))*-e^-t}{Sin(t)+Cos(t))*-e^-t}$

Edit: I worked the required expression backwards and got the necessary clue which resulted in this:

$$ \frac{\frac{\sqrt2}{2}(\sin t -\cos t)}{\frac{\sqrt2}{2}(\cos t +\sin t)}$$ (thanks Pauly B)

Answer 1

$$\begin{align} \frac{dy}{dx}&=\frac{\sin t-\cos t}{\cos t+\sin t}\\ &=\frac{\sqrt2(\sin t \cos\frac\pi4-\cos t\sin\frac\pi4)}{\sqrt2(\cos t \cos\frac\pi4+\sin t\sin\frac\pi4)}\\ &=\frac{\sin(t-\frac\pi4)}{\cos(t-\frac\pi4)}\\ &=\tan(t-\frac\pi4) \end{align}$$

Answer 2

$\textbf{Hint}$: $cos(t)^2+\sin(t)^2=1$, do $x^2+y^2$ and you will get a function $y$ who depends of $x$, then apply chain rule and dont forget the term $dx/dt$

Answer 3

Work backwards from the desired endpoint. There's really only one thing to be done with $\tan(t-\frac\pi4)$, and that's applying the addition formula: $$ \tan(t-\tfrac\pi4) = \frac{\tan(t)-\tan(\tfrac\pi4)}{1+\tan(t)\tan(\tfrac\pi4)} = \frac{\tan(t)-1}{1+\tan(t)} $$ What could you do next with this, to make it look more like $\frac{\sin(t)-\cos(t)}{\sin(t)+\cos(t)}$?