Find the equations of the bisectors of the angles between the pair of straight lines given by $\;$ $x^2 + 2xy \cot \alpha + y^2 = 0$
Equation of given pair of lines: $\;$ $x^2 + 2xy \cot \alpha + y^2 = 0$ $\;\;\; \cdots \; (1)$
Comparing with the standard equation of pair of lines: $\;$ $ax^2 + 2hxy + by^2 = 0$ $\cdots \; (2)$ gives
$a = 1, \;\; h = \cot \alpha, \;\; b = 1$
The combined equation of the bisectors of equation $(2)$ is $\;\;$ $\dfrac{x^2 - y^2}{a - b} = \dfrac{xy}{h}$ $\;$
i.e. $\;$ $x^2 - y^2 = \left(a - b\right) \dfrac{xy}{h}$
$\therefore \;$ The combined equation of the bisectors of equation $(1)$ is
$x^2 - y^2 = \left(1 - 1\right) \times \dfrac{xy}{\cot \alpha}$
i.e. $\;$ $x^2 - y^2 = 0$