If the angle 2 theta is acute,then the acute | vedclass

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(D) The given equation is $x^2(\cos 	heta - \sin 	heta) + 2xy \cos 	heta + y^2(\cos 	heta + \sin 	heta) = 0$.Comparing this with the general form

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$	heta$

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(D) The given equation is $x^2(\cos 	heta - \sin 	heta) + 2xy \cos 	heta + y^2(\cos 	heta + \sin 	heta) = 0$.Comparing this with the general form $ax^2 + 2hxy + by^2 = 0$,we get:$a = \cos 	heta - \sin 	heta$,$h = \cos 	heta$,and $b = \cos 	heta + \sin 	heta$.The formula for the acute angle $\alpha$ between the pair of lines is given by $	an \alpha = \left| \frac{2 \sqrt{h^2 - ab}}{a + b} ight|$.Substituting the values:$h^2 - ab = \cos^2 	heta - (\cos 	heta - \sin 	heta)(\cos 	heta + \sin 	heta) = \cos^2 	heta - (\cos^2 	heta - \sin^2 	heta) = \sin^2 	heta$.$a + b = (\cos 	heta - \sin 	heta) + (\cos 	heta + \sin 	heta) = 2 \cos 	heta$.Thus,$	an \alpha = \left| \frac{2 \sqrt{\sin^2 	heta}}{2 \cos 	heta} ight| = \left| \frac{2 \sin 	heta}{2 \cos 	heta} ight| = |	an 	heta|$.Since $2 	heta$ is acute,$	heta$ is also acute,so $\alpha = 	heta$.

If the angle $2 \theta$ is acute,then the acute angle between the pair of straight lines $x^2(\cos \theta - \sin \theta) + 2xy \cos \theta + y^2(\cos \theta + \sin \theta) = 0$ is:

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