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

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(B) The given equation 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

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$\frac{\pi}{4}$

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(B) The given equation is $(x^2+y^2) \cos^2 	heta = (x \cos 	heta + y \sin 	heta)^2$. Expanding the right side: $(x^2+y^2) \cos^2 	heta = x^2 \cos^2 	heta + y^2 \sin^2 	heta + 2xy \sin 	heta \cos 	heta$. Rearranging the terms to the form $Ax^2 + 2Hxy + By^2 = 0$: $x^2(\cos^2 	heta - \cos^2 	heta) + 2xy(\sin 	heta \cos 	heta) + y^2(\sin^2 	heta - \cos^2 	heta) = 0$. $0x^2 + 2xy(\sin 	heta \cos 	heta) - y^2(\cos^2 	heta - \sin^2 	heta) = 0$. For a pair of lines to be perpendicular,the sum of the coefficients of $x^2$ and $y^2$ must be zero $(A + B = 0)$. Here,$A = 0$ and $B = -(\cos^2 	heta - \sin^2 	heta) = -\cos(2	heta)$. Setting $A + B = 0$: $0 - \cos(2	heta) = 0 \Rightarrow \cos(2	heta) = 0$. Thus,$2	heta = \frac{\pi}{2} \Rightarrow 	heta = \frac{\pi}{4}$.

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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