If the pair of lines given by (x cos alpha + | vedclass

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

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$\frac{\pi}{4}$

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(C) Given equation: $(x \cos \alpha + y \sin \alpha)^2 = (x^2 + y^2) \sin^2 \alpha$ Expanding the left side: $x^2 \cos^2 \alpha + y^2 \sin^2 \alpha + 2xy \sin \alpha \cos \alpha = x^2 \sin^2 \alpha + y^2 \sin^2 \alpha$ Subtracting $y^2 \sin^2 \alpha$ from both sides: $x^2 \cos^2 \alpha + 2xy \sin \alpha \cos \alpha = x^2 \sin^2 \alpha$ Rearranging: $x^2(\cos^2 \alpha - \sin^2 \alpha) + 2xy \sin \alpha \cos \alpha = 0$ This is a homogeneous equation of the form $ax^2 + 2hxy + by^2 = 0$,where $a = \cos^2 \alpha - \sin^2 \alpha$,$h = \sin \alpha \cos \alpha$,and $b = 0$. For the lines to be perpendicular,the condition is $a + b = 0$. Substituting the values: $(\cos^2 \alpha - \sin^2 \alpha) + 0 = 0$ $\cos^2 \alpha = \sin^2 \alpha$ $	an^2 \alpha = 1$ Since $\alpha$ is typically in the first quadrant for such problems,$\alpha = \frac{\pi}{4}$.

If the pair of lines given by $(x \cos \alpha + y \sin \alpha)^2 = (x^2 + y^2) \sin^2 \alpha$ are perpendicular to each other,then $\alpha$ is

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