If the lines given by (x^2+y^2) sin^2 alpha = | vedclass

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(C) The given equation is $(x^2+y^2) \sin^2 \alpha = (x \cos \alpha - y \sin \alpha)^2$. Expanding the right side: $(x^2+y^2) \sin^2 \alpha = x^2 \cos

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$\frac{3}{2}$

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(C) The given equation is $(x^2+y^2) \sin^2 \alpha = (x \cos \alpha - y \sin \alpha)^2$. 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 the terms: $x^2(\sin^2 \alpha - \cos^2 \alpha) + 2xy \sin \alpha \cos \alpha + y^2(\sin^2 \alpha - \sin^2 \alpha) = 0$. This simplifies to: $-x^2 \cos 2\alpha + xy \sin 2\alpha = 0$. For a pair of lines $Ax^2 + 2Hxy + By^2 = 0$ to be perpendicular,the condition is $A + B = 0$. Here,$A = -\cos 2\alpha$ and $B = 0$. Thus,$-\cos 2\alpha + 0 = 0$,which implies $\cos 2\alpha = 0$. Since $\cos 2\alpha = 1 - 2\sin^2 \alpha = 0$,we get $\sin^2 \alpha = \frac{1}{2}$. Then $\cos^2 \alpha = 1 - \frac{1}{2} = \frac{1}{2}$. Therefore,$	an^2 \alpha = \frac{\sin^2 \alpha}{\cos^2 \alpha} = \frac{1/2}{1/2} = 1$. Finally,$\sin^2 \alpha + 	an^2 \alpha = \frac{1}{2} + 1 = \frac{3}{2}$.

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

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