Question: Find the angle between the lines represented by the following equation: x² - 2xy cot \alpha - y² = 0.

Solution:
Given,

Single equation of two straight lines is 
x²- 2xy\; cot \alpha - y²=0 - (i)

Comparing equation (i) with ax² + 2hxy + by²= 0, we get,
h = - cot \alpha, a = 1, b = -1

Angle represented by the following single equation is given by:

tan \theta = \left ( \pm \dfrac{2 \sqrt{h^2 - ab}}{a +b} \right )

or, tan \theta = \left ( \pm \dfrac{2 \sqrt{ (- cot \alpha)^2 - 1×(-1)}}{1 -1} \right )

or, tan \theta = \left ( \pm \dfrac{2 \sqrt{cot^2 \alpha +1}}{0} \right )

or, tan \theta = \pm 0

or, tan \theta = tan 90°

\theta = 90°

Therefore, the required angle between the lines represented by the given single equation is 90°.