Find the acute angle between the pair of straight lines $\;$ $x^2 + 2xy \sec x + y^2 = 0$
Equation of given pair of lines: $\;$ $x^2 + 2xy \sec x + y^2 = 0$
Comparing with the standard equation of pair of lines: $\;$ $ax^2 + 2hxy + by^2 = 0$ $\;$ gives
$a = 1, \;\; h = \sec x, \;\; b = 1$
Let $\theta$ be the angle between the given pair of lines.
Then, $\;$ $\tan \theta = \left|\dfrac{2 \sqrt{h^2 - ab}}{a + b}\right|$
i.e. $\;$ $\tan \theta = \left|\dfrac{2 \sqrt{\sec^2 x - 1 \times 1}}{1 + 1}\right|$
i.e. $\;$ $\tan \theta = \left|\sqrt{\sec^2 x - 1}\right| = \sqrt{\tan^2 x} = \tan x$
$\implies$ $\theta = \tan^{-1} \left(\tan x\right) = x$