For α ∈ 0 , π 2 , the angle between the lines represented by x cos θ - y cos θ +…

Given,

$\left[x\cos \theta -y\right]\left[\left(\cos \theta +\tan \alpha \right)x-\left(1-\cos \theta ·\tan \alpha \right)y\right]=0$

Now taking line $L_1\to x\cos \theta -y=0$

And $L_2\to \left(\cos \theta +\tan \alpha \right)x-\left(1-\cos \theta ·\tan \alpha \right)y=0$

Now slope of line $L_1$ is $m_1=\cos \theta$ and slope of line $L_2$ is $m_2=\frac{\cos \theta +\tan \alpha }{1-\cos \theta .\tan \alpha }$

Now we know that angle between two line is given by $\tan \varphi =\left|\frac{m_2-m_1}{1+m_1{m}_2}\right|$

$\Rightarrow \tan \phi =\left|\frac{\frac{\cos \theta +\tan \alpha }{1-\cos \theta .\tan \alpha }-\cos \theta }{1+\frac{\cos \theta +\tan \alpha }{1-\cos \theta .\tan \alpha }\times \cos \theta }\right|$

$\Rightarrow \tan \phi =\left|\frac{\cos \theta +\tan \alpha -\cos \theta +{\cos }^2\theta .\tan \alpha }{1-\cos \theta .\tan \alpha +{\cos }^2\theta +\tan \alpha .\cos \theta }\right|$

$\Rightarrow \tan \phi =\left|\frac{\tan \alpha \left(1+{\cos }^2\theta \right)}{1+{\cos }^2\theta }\right|$

$\Rightarrow \tan \phi =\tan \alpha$

$\Rightarrow \phi =\alpha$