If the pair of lines given by (x^2+y^2) cos^2 | vedclass

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(B) The given equation of the pair of lines is: $(x^2+y^2) \cos^2 	heta = (x \cos 	heta + y \sin 	heta)^2$ Expanding the right side: $(x^2+y^2) \co

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$\frac{\pi}{4}$

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(B) The given equation of the pair of lines is: $(x^2+y^2) \cos^2 	heta = (x \cos 	heta + y \sin 	heta)^2$ Expanding the right side: $(x^2+y^2) \cos^2 	heta = x^2 \cos^2 	heta + y^2 \sin^2 	heta + 2xy \sin 	heta \cos 	heta$ Subtracting $x^2 \cos^2 	heta$ from both sides: $y^2 \cos^2 	heta = y^2 \sin^2 	heta + 2xy \sin 	heta \cos 	heta$ Rearranging into the general form $Ax^2 + 2Hxy + By^2 = 0$: $0x^2 + (2 \sin 	heta \cos 	heta)xy + (\sin^2 	heta - \cos^2 	heta)y^2 = 0$ For the pair of lines to be perpendicular,the sum of the coefficients of $x^2$ and $y^2$ must be zero: $A + B = 0$ $0 + (\sin^2 	heta - \cos^2 	heta) = 0$ $\sin^2 	heta = \cos^2 	heta$ $	an^2 	heta = 1$ $	an 	heta = \pm 1$ Thus,$	heta = \frac{\pi}{4}$ or $\frac{3\pi}{4}$.

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

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