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(C) Given equation: $(x^2+y^2) \sin^2 \alpha = (x \cos \alpha - y \sin \alpha)^2$ Expanding the right side: $(x^2+y^2) \sin^2 \alpha = x^2 \cos^2 \alp

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$2\alpha$

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(C) Given equation: $(x^2+y^2) \sin^2 \alpha = (x \cos \alpha - y \sin \alpha)^2$ Expanding the right side: $(x^2+y^2) \sin^2 \alpha = x^2 \cos^2 \alpha + y^2 \sin^2 \alpha - 2xy \sin \alpha \cos \alpha$ Rearranging terms: $x^2 \sin^2 \alpha + y^2 \sin^2 \alpha = x^2 \cos^2 \alpha + y^2 \sin^2 \alpha - 2xy \sin \alpha \cos \alpha$ Simplifying: $x^2 \sin^2 \alpha = x^2 \cos^2 \alpha - 2xy \sin \alpha \cos \alpha$ $x^2(\sin^2 \alpha - \cos^2 \alpha) + 2xy \sin \alpha \cos \alpha = 0$ $-x^2 \cos(2\alpha) + xy \sin(2\alpha) = 0$ This is a homogeneous equation of the form $Ax^2 + 2Hxy + By^2 = 0$,where $A = -\cos(2\alpha)$,$H = \frac{1}{2} \sin(2\alpha)$,and $B = 0$. The angle $	heta$ between the lines is given by $	an 	heta = \left| \frac{2\sqrt{H^2 - AB}}{A+B} ight|$. Substituting values: $	an 	heta = \left| \frac{2\sqrt{\frac{1}{4} \sin^2(2\alpha) - 0}}{-\cos(2\alpha) + 0} ight| = \left| \frac{\sin(2\alpha)}{-\cos(2\alpha)} ight| = |-	an(2\alpha)| = |	an(2\alpha)|$. Thus,$	heta = 2\alpha$.

The angle between the straight lines represented by $(x^2+y^2) \sin^2 \alpha = (x \cos \alpha - y \sin \alpha)^2$ is

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