If theta is the angle between the lines x^2+2 | vedclass

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(A) The angle $	heta$ between the pair of lines $ax^2 + 2hxy + by^2 = 0$ is given by $	an 	heta = \left| \frac{2\sqrt{h^2 - ab}}{a+b} ight|$.For

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$	heta$

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(A) The angle $	heta$ between the pair of lines $ax^2 + 2hxy + by^2 = 0$ is given by $	an 	heta = \left| \frac{2\sqrt{h^2 - ab}}{a+b} ight|$.For the equation $x^2 + 2xy \sec 	heta + y^2 = 0$,we have $a = 1$,$h = \sec 	heta$,and $b = 1$.Let $\phi$ be the angle between these lines.Then $	an \phi = \left| \frac{2\sqrt{(\sec 	heta)^2 - (1)(1)}}{1+1} ight|$.$	an \phi = \left| \frac{2\sqrt{\sec^2 	heta - 1}}{2} ight|$.Since $\sec^2 	heta - 1 = 	an^2 	heta$,we have $	an \phi = \sqrt{	an^2 	heta} = 	an 	heta$.Therefore,$\phi = 	heta$.

If $\theta$ is the angle between the lines $x^2+2 h x y+b y^2=0$,then the angle between $x^2+2 x y \sec \theta+y^2=0$ is

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