The acute angle between the pair of lines $\left(\mathrm{cos\alpha }+\mathrm{sin\alpha }\right){\mathrm{x}}^2-2\mathrm{xycos\alpha }+\left(\mathrm{cos\alpha }-\mathrm{sin\alpha }\right){\mathrm{y}}^2=0$ is

If $\mathrm{\theta }$ is acute angle between pair of lines represented by the equation ${\mathrm{ax}}^2+2\mathrm{hxy}+{\mathrm{by}}^2=0$ then $\mathrm{cos\theta }=\frac{\left|\mathrm{a}+\mathrm{b}\right|}{\sqrt{{(\mathrm{a}-\mathrm{b})}^2+{\left(2\mathrm{h}\right)}^2}}$
For the equation $\left(\mathrm{cos\alpha }+\mathrm{sin\alpha }\right){\mathrm{x}}^2-2\mathrm{xycos\alpha }+\left(\mathrm{cos\alpha }-\mathrm{sin\alpha }\right){\mathrm{y}}^2=0$, the values are $\mathrm{a}=\mathrm{cos\alpha }+\mathrm{sin\alpha },2\mathrm{h}=-2\mathrm{cos\alpha },\mathrm{b}=\mathrm{cos\alpha }-\mathrm{sin\alpha }$
Hence, 
$\begin{array}{c}\mathrm{cos\theta }=\frac{2\mathrm{cos\alpha }}{\sqrt{4{\sin }^2\mathrm{\alpha }+4{\cos }^2\mathrm{\alpha }}}\\ =\frac{2\mathrm{cos\alpha }}{2}\\ =\mathrm{cos\alpha }\end{array}$
Therefore, $\mathrm{\theta }=\boxed{\mathrm{\alpha }}$