[tex]\bold {\frac{dy}{dx} = \frac{(\frac{dy}{dt})}{ (\frac{dx}{dt})}} [/tex]
Using product rule: f'g + fg'
[tex]\frac{dy}{dt} = -e^{-t} sin (t) + e^{-t} cos (t) \\ \\ \frac{dx}{dt} = -e^{-t} cos (t) - e^{-t} sin (t)[/tex]
Factor out [tex]-e^{-t}[/tex]
[tex]\bold {\frac{dy}{dx} = \frac{sin (t) - cos(t)}{sin (t) + cos (t)}}[/tex]
Next, we have to use trig identities to equate this with tan(t - pi/4).
There are a lot of steps here, i think its easiest to start at either end and meet in the middle. I will show that:
[tex]tan(t - \frac{\pi}{4}) = tan(2t) - sec(2t) = \frac{sin (t) - cos(t)}{sin(t)+cos(t)}[/tex]
Starting on left side, use angle sum identity for tan. Note tan(pi/4) = 1.
[tex]tan(t - \frac{\pi}{4}) = \frac{tan(t) - 1}{1+tan(t)} \\ \\ = \frac{tan(t) - 1}{1+tan(t)}*\frac{1-tan(t)}{1-tan(t)} \\ \\ =\frac{2 tan(t) - (1+tan^2(t))}{1-tan^2(t)} \\ \\ = \frac{2 tan(t) - sec^2 (t))}{1-tan^2(t)} \\ \\ = tan(2t) - \frac{sec^2 (t))}{1-tan^2(t)} \\ \\ =tan(2t) - \frac{1}{cos^2(t) - sin^2(t)} \\ \\ = tan(2t) - sec(2t)[/tex]
Now starting on right side:
[tex]\frac{sin(t) - cos(t)}{sin(t) + cos(t)} \\ \\ =\frac{sin(t) - cos(t)}{sin(t) + cos(t)} * \frac{sin(t) - cos(t)}{sin(t)-cos(t)} \\ \\ = \frac{sin^2(t) + cos^2(t) -2sin(t)cos(t)}{sin^2(t) -cos^2(t)} \\ \\ = \frac{1-sin(2t)}{-cos(2t)} \\ \\ =tan(2t) - sec(2t)[/tex]
Therefore proving that
[tex]\bold{\frac{dy}{dx} = \frac{sin(t) - cos(t)}{sin(t) + cos(t)} = tan(t - \frac{\pi}{4})}[/tex]
