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(A) The given equation of the pair of lines is $x^2+\lambda xy-y^2 	an^2 	heta=0$.Comparing this with the general form $ax^2+2hxy+by^2=0$,we get $a=

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(A) The given equation of the pair of lines is $x^2+\lambda xy-y^2 	an^2 	heta=0$.Comparing this with the general form $ax^2+2hxy+by^2=0$,we get $a=1$,$h=\frac{\lambda}{2}$,and $b=-	an^2 	heta$.The angle $\alpha$ between the pair of lines is given by $	an \alpha = \left|\frac{2\sqrt{h^2-ab}}{a+b}ight|$.Here,$\alpha = 2	heta$,so $	an 2	heta = \left|\frac{2\sqrt{(\frac{\lambda}{2})^2 - (1)(-	an^2 	heta)}}{1-	an^2 	heta}ight|$.Using the identity $	an 2	heta = \frac{2	an 	heta}{1-	an^2 	heta}$,we have $\frac{2	an 	heta}{1-	an^2 	heta} = \left|\frac{2\sqrt{\frac{\lambda^2}{4}+	an^2 	heta}}{1-	an^2 	heta}ight|$.Squaring both sides,we get $\frac{4	an^2 	heta}{(1-	an^2 	heta)^2} = \frac{4(\frac{\lambda^2}{4}+	an^2 	heta)}{(1-	an^2 	heta)^2}$.This simplifies to $	an^2 	heta = \frac{\lambda^2}{4} + 	an^2 	heta$,which implies $\frac{\lambda^2}{4} = 0$.Therefore,$\lambda = 0$.

If the angle between the lines represented by the equation $x^2+\lambda xy-y^2 \tan^2 \theta=0$ is $2\theta$,then the value of $\lambda$ is

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