What is the angle between the lines given by ${x^2} - {y^2} = 0$ ?
A. ${15^ \circ }$
B. ${45^ \circ }$
C. ${75^ \circ }$
D. ${90^ \circ }$

Hint: A pair of straight lines, passing through the origin, are represented by a general equation of the form $a{x^2} + 2hxy + b{y^2}=0$ . Sum of the slopes of the two lines is given by $\dfrac{{ - 2h}}{b}$ and the product of the slopes is given by $\dfrac{a}{b}$ . The angle between the two lines, $\theta $ , is calculated using the formula $\tan \theta = \left| {\dfrac{{2\sqrt {{h^2} - ab} }}{{a + b}}} \right|$. We will use this formula to derive the condition and use it to get the desired solution.

Formula Used: The angle between a pair of straight lines, passing through the origin, represented by $a{x^2} + 2hxy + b{y^2}=0$ is calculated using the formula $\tan \theta = \left| {\dfrac{{2\sqrt {{h^2} - ab} }}{{a + b}}} \right|$ .

Complete step-by-step solution:
Given equation of two straight lines:
${x^2} - {y^2} = 0$ … (1)
General form of a pair of straight lines passing through the origin:
$a{x^2} + 2hxy + b{y^2}=0$ … (2)
Comparing equation (1) with the general form given in equation (2),
$a = 1$ ,
$b = - 1$ and
$h = 0$
Now, we know that the tangent of the angle, let’s say $\theta $ , between two lines is calculated using:
$\tan \theta = \left| {\dfrac{{2\sqrt {{h^2} - ab} }}{{a + b}}} \right|$
Substituting the values of the variables, we get:
$\tan \theta = \left| {\dfrac{{2\sqrt {0 + 1} }}{{1 + ( - 1)}}} \right|$
On simplifying further, we get $\tan \theta = \infty $ .
Calculating the inverse, we get $\theta = \dfrac{\pi }{2}$ .
Hence, the angle between the given lines is ${90^ \circ }$ .
Thus, the correct option is D.

Note: Compare the given equation with the general form of a pair of straight lines to get the correct values of the coefficients. Then substitute these values in the formula, to calculate the tangent of the angle between the pair of straight lines, correctly to avoid any miscalculations. If students remember the perpendicular lines condition then it is easy to answer this question.