If x=3 cosθ-2cos3θ, y=3 sinθ-2 sin3θ then dydx=

Given $$x=3cos$$θ$$-2cos3$$θ differentiate w.r.to x on both sides dxdθ$$=-3sin$$θ$$-2(-3cos2$$θ$$sin$$θ$$)$$ ⇒dxdθ $$=-3sin$$θ$$+6cos2$$θ$$sin$$θ $$Giveny=3sin$$θ$$-2sin3$$θ differentiate w.r.to x on both sides dydθ$$=3cos$$θ$$-2(3sin2$$θ$$cos$$θ$$)$$ ⇒dydθ$$=3cos$$θ$$-6sin2$$θ$$cos$$θ Now $$dydx=dyd$$θdxdθ $$dydx=3cos$$θ$$-6$$ $$sin2$$θ$$cos$$θ$$-3sin$$θ$$+6$$ $$cos2$$θ$$sin$$θ $$=3cos$$θ(1-2$$ $$sin2$$θ$$)3sin$$θ$$(-1+2$$ $$cos2$$θ$$) $$=$$$$cos$$θ$$cos2$$θ$$sin$$θ.$$cos2$$θ $$=$$$$cot$$θ

Q.

If $x = 3 cosθ - 2 \cos^{3} θ, y = 3 sinθ - 2 \sin^{3} θ then \frac{dy}{dx} =$

Moderate

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$\tan θ$

$\cot θ$

$\cot \frac{θ}{2}$

$\tan \frac{θ}{2}$

answer is B.

(Detailed Solution Below)

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Detailed Solution

$G i v e n x = 3 \cos θ - 2 \cos^{3} θ d i f f e r e n t i a t e w . r . t o x o n b o t h s i d e s \frac{d x}{d θ} = - 3 \sin θ - 2 (- 3 \cos^{2} θ \sin θ) \Rightarrow \frac{d x}{d θ} = - 3 \sin θ + 6 \cos^{2} θ \sin θ G i v e n y = 3 \sin θ - 2 \sin^{3} θ d i f f e r e n t i a t e w . r . t o x o n b o t h s i d e s \frac{d y}{d θ} = 3 \cos θ - 2 (3 \sin^{2} θ \cos θ) \Rightarrow \frac{d y}{d θ} = 3 \cos θ - 6 \sin^{2} θ \cos θ N o w \frac{dy}{dx} = \frac{\frac{d y}{d θ}}{\frac{d x}{d θ}} \frac{dy}{dx} = \frac{3 cosθ - 6 \sin^{2} θcosθ}{- 3 \sin θ + 6 \cos^{2} θ \sin θ} = \frac{3 cosθ (1 - 2 \sin^{2} θ)}{3 \sin θ (- 1 + 2 \cos^{2} θ)} = \frac{cosθcos 2 θ}{sinθ . \cos 2 θ} = c o t θ$

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