Parametric differentiation

Question

A question from Active maths. At 60 years old this is my interest not my homework!!

Let $$\begin{cases}y=e^t\cos t\\ x=e^t\sin t,\end{cases}$$

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

I have worked out that $dy/dx = \cos t-\sin t/\sin t+\cos t$, but I cannot see how to turn this into $\tan(\pi/4 -t)$.

I am happy in the first instance to have a hint or two.
Thanks

Answer 1

Divide top and bottom of your fraction by $\cos t$ to get

$$(*)\quad \frac{1 - \frac{\sin t}{\cos t}}{\frac{\sin t}{\cos t} + 1} = \frac{1 - \tan t}{\tan t + 1}.$$

Now use the angle subtraction formula

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

to write the right-hand side of $(*)$ as $\tan(\pi/4 - t)$.

Answer 2

Hint: Note that you can write $$\cos t - \sin t = \sqrt 2 \left(\tfrac1{\sqrt 2}\cos t - \tfrac1{\sqrt 2}\sin t\right) = \sqrt 2 \left(\sin\tfrac{\pi}4\cos t - \cos\tfrac{\pi}4\sin t\right)$$ and $$\cos t + \sin t = \sqrt 2 \left(\tfrac1{\sqrt 2}\cos t + \tfrac1{\sqrt 2}\sin t\right) = \sqrt 2 \left(\cos \tfrac{\pi}4\cos t + \sin\tfrac{\pi}4\sin t\right)$$ Now recall the angle sum (difference) formulas.

Answer 3

$$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{-e^t \sin t + e^t \cos t}{e^t \cos t + e^t \sin t} = \frac{\cos t - \sin t}{\cos t + \sin t} = \frac{1 - \tan t}{1 + \tan t} = \frac{tan \frac{\pi}{4} - \tan t}{\tan \frac{\pi}{4} + \tan t} = \tan \left( \frac{\pi}{4} - t \right),$$ where we employed the product rule for differentiation of $x$ and $y$ with respect to $t$, and the tangent angle addition identity $$\tan (a+b) = \frac{\tan a + \tan b}{1 - \tan a \tan b}$$ with the choice $a = \pi/4$, $b = -t$.