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The parametric equations of a curve are x = e^−t cos t y = e^−t sin t. Show that dy/dx= tan (t −1/4π)
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@mathmale
I differentiate both side with respect to t and got \[dy/dx = \frac{ e ^{-t }\cos t-e ^{-t}\sin t }{ -e ^{-t}\sin t-e ^{-t}\cos t }\]
@ganeshie8 please help
\[\frac{dy}{dx} = \frac{ e ^{-t }\cos t-e ^{-t}\sin t }{ -e ^{-t}\sin t-e ^{-t}\cos t }=\frac{\sin t-\cos t }{ \sin t+\cos t }\] \[\begin{align*} \frac{\sin t-\cos t}{\sin t+\cos t}&=\frac{\dfrac{\sin t}{\cos t}-\dfrac{\cos t}{\cos t}}{\dfrac{\sin t}{\cos t}+\dfrac{\cos t}{\cos t}}\\\\ &=\frac{\tan t-1}{1+\tan t}\\\\ &=\frac{\tan t-\tan\dfrac{\pi}{4}}{1+\tan t\tan\dfrac{\pi}{4}}\\\\ &=\tan\left(t-\frac{\pi}{4}\right) \end{align*}\]
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