If frac{y}{x} cos^4 alpha + frac{x}{y} sin^4 | vedclass

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Question

(C) Given the equation: $\frac{y}{x} \cos^4 \alpha + \frac{x}{y} \sin^4 \alpha = 2 \sin^2 \alpha \cos^2 \alpha$ Multiply both sides by $xy$: $y^2 \cos

Vedclass · Accepted Answer

$\frac{\sin^2 \alpha}{\cos^2 \alpha}$

Vedclass · Answer

(C) Given the equation: $\frac{y}{x} \cos^4 \alpha + \frac{x}{y} \sin^4 \alpha = 2 \sin^2 \alpha \cos^2 \alpha$ Multiply both sides by $xy$: $y^2 \cos^4 \alpha + x^2 \sin^4 \alpha = 2xy \sin^2 \alpha \cos^2 \alpha$ Rearrange the terms: $x^2 \sin^4 \alpha - 2xy \sin^2 \alpha \cos^2 \alpha + y^2 \cos^4 \alpha = 0$ This is a perfect square: $(x \sin^2 \alpha - y \cos^2 \alpha)^2 = 0$ Taking the square root on both sides: $x \sin^2 \alpha - y \cos^2 \alpha = 0$ $y \cos^2 \alpha = x \sin^2 \alpha$ $y = x \frac{\sin^2 \alpha}{\cos^2 \alpha}$ Differentiating with respect to $x$: $\frac{dy}{dx} = \frac{\sin^2 \alpha}{\cos^2 \alpha}$

If $\frac{y}{x} \cos^4 \alpha + \frac{x}{y} \sin^4 \alpha = 2 \sin^2 \alpha \cos^2 \alpha$,then $\frac{dy}{dx} = $

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