If x = acos4θ, y = asin4θ show that \(\frac{dy}{dx}\) = -tan2θ

1.	If x = acos4θ, y = asin4θ show that \(\frac{dy}{dx}\) = -tan2θ
Answer» Given x = a cos⁴θ, y = a sin 4θ \(\frac{dx}{d\theta}\) = a(4cos³θ( – sinθ)) \(\frac{dy}{d\theta}\) = 4a sin³θcosθ ∴ \(\frac{dy}{dx}\) = \(\frac{4asin^3\theta cos\theta}{-4acos^3\theta sin\theta}\) = - tan²θ

Answer»

Given x = a cos⁴θ, y = a sin 4θ

\(\frac{dx}{d\theta}\) = a(4cos³θ( – sinθ)) \(\frac{dy}{d\theta}\) = 4a sin³θcosθ

∴ \(\frac{dy}{dx}\) = \(\frac{4asin^3\theta cos\theta}{-4acos^3\theta sin\theta}\) = - tan²θ

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