Question

$x=e^{\theta }(\sin \theta +\cos \theta )$, $y=e^{\theta }(\sin \theta -\cos \theta )$

Fine $\frac{dy}{dx}$.

Answer 1

Consider the following equation.

$x=e^{\theta }(sin\theta +cos\theta ),y=e^{\theta }(sin\theta -cos\theta )$

Differentiate $x\&y$ w.r.t $\theta$

$x\&y$

$\frac{dx}{d\theta }=e^{\theta }(sin\theta +cos\theta )+e^{\theta }(cos\theta -sin\theta ).......\left(1\right)$

$\frac{dy}{d\theta }=e^{\theta }(sin\theta -cos\theta )+e^{\theta }(cos\theta +sin\theta ).......\left(2\right)$

Divide by eq..(2) In eq. (1)

$\frac{dy}{dx}=\frac{e^{\theta }(sin\theta -cos\theta +cos\theta +sin\theta )}{e^{\theta }(sin\theta +cos\theta +cos\theta -sin\theta )}$

$=\frac{2\sin \theta }{2cos\theta }$

$=\tan \theta$

Hence, this is the correct answer.