Question

If $x=2\cos t-\cos 2t$ and $y=2\sin t-\sin 2t,$ then $\frac{dy}{dx}$ is equal to

• A. $\tan t$
• B. $-\tan \frac{t}{2}$
• C. $\tan \left(\frac{3t}{2}\right)$
• D. $-\tan \left(\frac{3t}{2}\right)$

Answer 1

The correct option is C $\tan \left(\frac{3t}{2}\right)$

$x=2\cos t-\cos 2t$
$\Rightarrow \frac{dx}{dt}=-2\sin t+2\sin 2t$
$\Rightarrow \frac{dx}{dt}=2(\sin 2t-\sin t)$
$y=2\sin t-\sin 2t$
$\Rightarrow \frac{dy}{dt}=2(\cos t-\cos 2t)$

Now,
$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{\cos t-\cos 2t}{\sin 2t-\sin t}=\frac{2\sin \frac{3t}{2}\cdot \sin \frac{t}{2}}{2\cos \frac{3t}{2}\cdot \sin \frac{t}{2}}$
$=\tan \left(\frac{3t}{2}\right)$