Q62 / 150 · MHT CET 2009 · Mathematics

Find $\frac{d y}{d x}$, if $x=2 \cos \theta-\cos 2 \theta$ and $y=2 \sin \theta-\sin 2 \theta .$

  • Correct answer
    A
    $\tan \frac{3 \theta}{2}$
  • B
    $-\tan \frac{3 \theta}{2}$
  • C
    $\cot \frac{3 \theta}{2}$
  • D
    $-\cot \frac{3 \theta}{2}$

Solution

Given, $x=2 \cos \theta-\cos 2 \theta$ and $y=2 \sin \theta-\sin 2 \theta$ $$ \begin{aligned} \frac{d x}{d \theta} &=-2 \sin \theta+2 \sin 2 \theta \\ \text { and } \frac{d y}{d \theta} &=2 \cos \theta-2 \cos 2 \theta \\ \therefore \quad \frac{d y}{d x} &=\frac{2 \cos \theta-2 \cos 2 \theta}{-2 \sin \theta+2 \sin 2 \theta} \\ &=\frac{\cos \theta-\cos 2 \theta}{\sin 2 \theta-\sin \theta} \\ &=\frac{2 \sin \left(\frac{\theta+2 \theta}{2}\right) \sin \left(\frac{2 \theta-\theta}{2}\right)}{2 \cos \left(\frac{\theta+2 \theta}{2}\right) \sin \left(\frac{2 \theta-\theta}{2}\right)} \\ &=\tan \frac{3 \theta}{2} \end{aligned} $$

Hence, the correct answer is option: 'A'.

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