$Theanglebetweenthepairoflines{y}^2-2xy\cos ec\theta +x^2=0,0\le \theta \le \frac{\pi }{2},is$

$\begin{array}{l}GivenEquation{x}^2-2xy\cos ce\theta +y^2=0\\ Comparingwitha{x}^2+2hxy+b{y}^2=0\\ a=1,2h=-2\cos ce\theta ,b=1\\ Let\text{'}\alpha \text{'}betheangleb/wthepairoflines\\ \therefore \cos \alpha =\frac{a+b}{\sqrt{{\left(a-b\right)}^2+4h^2}}=\frac{1+1}{\sqrt{{\left(1-1\right)}^2+4\cos c{e}^2\theta }}=\frac{2}{2\cos ce\theta }=\sin \theta =\cos \left(\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.-\theta \right)\\ \therefore \alpha =\frac{\pi }{2}-\theta \end{array}$