For alpha in [0, frac{pi}{2}],the angle betwe | vedclass

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(A) The given equation is $[x \cos 	heta - y][(\cos 	heta + 	an \alpha) x - (1 - \cos 	heta 	an \alpha) y] = 0$.This represents two lines:$L_1: x

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$\alpha$

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(A) The given equation is $[x \cos 	heta - y][(\cos 	heta + 	an \alpha) x - (1 - \cos 	heta 	an \alpha) y] = 0$.This represents two lines:$L_1: x \cos 	heta - y = 0 \implies y = (\cos 	heta) x$,so slope $m_1 = \cos 	heta$.$L_2: (\cos 	heta + 	an \alpha) x - (1 - \cos 	heta 	an \alpha) y = 0 \implies y = \frac{\cos 	heta + 	an \alpha}{1 - \cos 	heta 	an \alpha} x$,so slope $m_2 = \frac{\cos 	heta + 	an \alpha}{1 - \cos 	heta 	an \alpha}$.The angle $\phi$ between two lines is given by $	an \phi = \left| \frac{m_2 - m_1}{1 + m_1 m_2} ight|$.Substituting the values of $m_1$ and $m_2$:$	an \phi = \left| \frac{\frac{\cos 	heta + 	an \alpha}{1 - \cos 	heta 	an \alpha} - \cos 	heta}{1 + \left( \frac{\cos 	heta + 	an \alpha}{1 - \cos 	heta 	an \alpha} ight) \cos 	heta} ight|$$	an \phi = \left| \frac{\cos 	heta + 	an \alpha - \cos 	heta (1 - \cos 	heta 	an \alpha)}{1 - \cos 	heta 	an \alpha + \cos 	heta (\cos 	heta + 	an \alpha)} ight|$$	an \phi = \left| \frac{\cos 	heta + 	an \alpha - \cos 	heta + \cos^2 	heta 	an \alpha}{1 - \cos 	heta 	an \alpha + \cos^2 	heta + \cos 	heta 	an \alpha} ight|$$	an \phi = \left| \frac{	an \alpha (1 + \cos^2 	heta)}{1 + \cos^2 	heta} ight| = 	an \alpha$.Thus,$\phi = \alpha$.

For $\alpha \in [0, \frac{\pi}{2}]$,the angle between the lines represented by $[x \cos \theta - y][(\cos \theta + \tan \alpha) x - (1 - \cos \theta \tan \alpha) y] = 0$ is

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