MHT CET · Maths · Differentiation
Find \(\frac{d y}{d x}\), if \(x=2 \cos \theta-\cos 2 \theta\) and
\(y=2 \sin \theta-\sin 2 \theta .\)
- A \(\tan \frac{3 \theta}{2}\)
- B \(-\tan \frac{3 \theta}{2}\)
- C \(\cot \frac{3 \theta}{2}\)
- D \(-\cot \frac{3 \theta}{2}\)
Answer & Solution
Correct Answer
(A) \(\tan \frac{3 \theta}{2}\)
Step-by-step Solution
Detailed explanation
Given, \(x=2 \cos \theta-\cos 2 \theta\)
and \(y=2 \sin \theta-\sin 2 \theta\)
\(\frac{d x}{d \theta} =-2 \sin \theta+2 \sin 2 \theta \)
\( \text { and } \frac{d y}{d \theta} =2 \cos \theta-2 \cos 2 \theta \)
\( \therefore \frac{d y}{d x} =\frac{2 \cos \theta-2 \cos 2 \theta}{-2 \sin \theta+2 \sin 2 \theta} \)
\( =\frac{\cos \theta-\cos 2 \theta}{\sin 2 \theta-\sin \theta} \)
\( =\frac{2 \sin \left(\frac{\theta+2 \theta}{2}\right) \sin \left(\frac{2 \theta-\theta}{2}\right)}{2 \cos \left(\frac{\theta+2 \theta}{2}\right) \sin \left(\frac{2 \theta-\theta}{2}\right)} \)
\( =\tan \frac{3 \theta}{2}\)
and \(y=2 \sin \theta-\sin 2 \theta\)
\(\frac{d x}{d \theta} =-2 \sin \theta+2 \sin 2 \theta \)
\( \text { and } \frac{d y}{d \theta} =2 \cos \theta-2 \cos 2 \theta \)
\( \therefore \frac{d y}{d x} =\frac{2 \cos \theta-2 \cos 2 \theta}{-2 \sin \theta+2 \sin 2 \theta} \)
\( =\frac{\cos \theta-\cos 2 \theta}{\sin 2 \theta-\sin \theta} \)
\( =\frac{2 \sin \left(\frac{\theta+2 \theta}{2}\right) \sin \left(\frac{2 \theta-\theta}{2}\right)}{2 \cos \left(\frac{\theta+2 \theta}{2}\right) \sin \left(\frac{2 \theta-\theta}{2}\right)} \)
\( =\tan \frac{3 \theta}{2}\)
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