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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 $\tan\left(t-\dfrac{\pi}{4}\right)$

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    $\begingroup$ Do you know the identity for $\tan(A-B)$? Try that on $\tan(t-\pi/4)$ and simplify it to the other expression. $\endgroup$ – Rory Daulton Apr 7 '15 at 23:37
  • $\begingroup$ Recall the angle difference identities: $$\sin(x\pm y)=\sin x\cos y\pm \cos x\sin y\\ \cos(x\pm y)=\cos x\cos y\mp\sin x\sin y$$ $\endgroup$ – user170231 Apr 7 '15 at 23:38
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$$\frac{\sin t-\cos t}{\cos t+\sin t}=\frac{\sqrt 2(\sin t\,\cos \frac\pi 4-\sin\frac\pi 4\,\cos t)}{\sqrt 2(\cos t\,\cos \frac\pi 4+\sin t\,\sin\frac\pi 4)}=\frac{\sin(t-\frac\pi 4)}{\cos(t-\frac\pi 4)}.$$

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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.

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