Question

If $x=3\cos \theta -\cos 3\theta$ and $y=3\sin \theta -\sin 3\theta ,$ then $\frac{dy}{dx}$ is

• A. $\tan 2\theta$
• B. $\sin 2\theta$
• C. $-\tan 2\theta$
• D. $\cot 2\theta$

Answer 1

The correct option is A $\tan 2\theta$

We have,
$\frac{dx}{d\theta }=-3\sin \theta +3\sin 3\theta$
$\frac{dy}{d\theta }=3\cos \theta -3\cos 3\theta$
$\Rightarrow \frac{dy}{dx}=\frac{3\cos \theta -3\cos 3\theta }{-3\sin \theta +3\sin 3\theta }$
$=\frac{\cos \theta -\cos 3\theta }{\sin 3\theta -\sin \theta }\cdots \left(1\right)$
Since, $\cos C-\cos D=2\sin \left(\frac{C+D}{2}\right)\sin \left(\frac{D-C}{2}\right)$
and, $\sin C-\sin D=2\sin \left(\frac{C-D}{2}\right)\cos \left(\frac{C+D}{2}\right)$
equation $\left(1\right)$ can be written as $\frac{\sin 2\theta \sin \theta }{\cos 2\theta \sin \theta }=\tan 2\theta$