The angle between the straight lines represented by $\left(x^2+y^2\right){\sin }^2\alpha ={\left(x\cos \alpha -y\sin \alpha \right)}^{2}is$

Given line $\left(x^2+y^2\right){\sin }^2\alpha ={\left(x\cos \alpha -y\sin \alpha \right)}^2$

 $\begin{array}{l}{x}^2\left({\sin }^2\alpha -{\cos }^2\alpha \right)+2xy\sin \alpha \cos \alpha =0\\ \Rightarrow -x^2\cos 2\alpha +xy\sin 2\alpha =0\\ \Rightarrow x^2\cos 2\alpha -xy\sin 2\alpha =0\\ Comparingwitha{x}^2+2hxy+b{y}^2=0\\ a=\cos 2\alpha ,2h=-\sin 2\alpha ,b=0\\ Let\text{'}\theta \text{'}betheangleb/wthepairofline\\ \cos \theta =\frac{\left|\cos 2\alpha +0\right|}{\sqrt{{\left(\cos 2\alpha -0\right)}^2+{\left(-\sin 2\alpha \right)}^2}}\Rightarrow \frac{\cos 2\alpha }{\sqrt{{\cos }^{2}2\alpha +{\sin }^{2}2\alpha }}\\ \cos \theta =\cos 2\alpha \Rightarrow \theta =2\alpha \end{array}$