Let the equation ax² + 2hxy + by²= 0 repsresents a pair of straight lines. Then show that the angle $\mathrm{\theta }$ between the lines is given by $\cos \mathrm{\theta }=\pm \frac{\mathrm{a}+\mathrm{b}}{\sqrt{(\mathrm{a}-\mathrm{b}{)}^2+4{\mathrm{h}}^2}}$

$\text{ Let }{\mathrm{ax}}^2+2\mathrm{hxy}+{\mathrm{by}}^2=0\text{ represents two lines be }$
${\mathrm{l}}_1\mathrm{x}+{\mathrm{m}}_1\mathrm{y}=0 ...\left(1\right) \mathrm{and} {\mathrm{l}}_2\mathrm{x}+{\mathrm{m}}_2\mathrm{y}=0\dots \left(2\right)$
$\text{ Then }{\mathrm{l}}_1{\mathrm{l}}_2=\mathrm{a},{\mathrm{l}}_1{\mathrm{m}}_2+{\mathrm{l}}_2{\mathrm{m}}_1=2\mathrm{h},{\mathrm{m}}_1{\mathrm{m}}_2=\mathrm{b}$
$\text{ Let }\mathrm{\theta }\text{ be the angle between the lines (1) and (2) then }$
$\begin{array}{l}\cos \mathrm{\theta }=\pm \frac{{\mathrm{l}}_1{\mathrm{l}}_2+{\mathrm{m}}_1{\mathrm{m}}_2}{\sqrt{\left({\mathrm{l}}_1^2+{\mathrm{m}}_1^2\right)\left({\mathrm{l}}_2^2+{\mathrm{m}}_2^2\right)}}\\ =\pm \frac{{\mathrm{l}}_1{\mathrm{l}}_2+{\mathrm{m}}_1{\mathrm{m}}_2}{\sqrt{{\mathrm{l}}_1^2{\mathrm{l}}_2^2+{\mathrm{m}}_1^2{\mathrm{m}}_2^2+{\mathrm{l}}_1^2{\mathrm{m}}_2^2+{\mathrm{l}}_2^2{\mathrm{m}}_1^2}}\\ =\pm \frac{{\mathrm{l}}_1{\mathrm{l}}_2+{\mathrm{m}}_1{\mathrm{m}}_2}{\sqrt{{\left({\mathrm{l}}_1{\mathrm{l}}_2-{\mathrm{m}}_1{\mathrm{m}}_2\right)}^2+2{\mathrm{l}}_1{\mathrm{l}}_2{\mathrm{m}}_1{\mathrm{m}}_2+{\left({\mathrm{l}}_1{\mathrm{l}}_2+{\mathrm{m}}_1{\mathrm{m}}_2\right)}^2-2{\mathrm{l}}_1{\mathrm{l}}_2{\mathrm{m}}_1{\mathrm{m}}_2}}\\ \cos \mathrm{\theta }=\pm \frac{a+b}{\sqrt{(\mathrm{a}-\mathrm{b}{)}^2+4{\mathrm{h}}^2}}\end{array}$