If ${\rm{(a + 3b)(3a + b) = 4}}{{\rm{h}}^{\rm{2}}}$, then the angle between the lines represented by ${\rm{a}}{{\rm{x}}^{\rm{2}}}{\rm{ + 2hxy + b}}{{\rm{y}}^{\rm{2}}}{\rm{ = 0}}$ is
A. ${30^\circ }$
B. ${45^\circ }$
C. ${60^\circ }$
D. ${\tan ^{ - 1}}\left( {\dfrac{1}{2}} \right)$

Hint: Here we simplify the equation provided and put the values in the formula for the angle between a pair of straight lines and solve the equation further.

Formula Used:
${\rm{\tan\theta }} = \dfrac{{{\rm{2}}\sqrt {{\rm{(}}{{\rm{h}}^{\rm{2}}}{\rm{ - ab}})} }}{{(a + b)}}$
${a^2} + {b^2} = {(a + b)^2} - 2ab$

Complete step by step solution:
Here, we are given the equation
$(a + 3b)(3a + b) = 4{h^2}$
Multiplying the terms on the left side of the equation
$ \Rightarrow 3{a^2} + 9ab + ab + 3{b^2} = 4{h^2}$
Adding the terms
$ \Rightarrow 3{a^2} + 10ab + 3{b^2} = 4{h^2}$
Further as we know the angle between pair of straight lines $a{x^2} + 2hxy + b{y^2} = 0$ will be ${\rm{tan\theta }} = \dfrac{{{\rm{2}}\sqrt {{\rm{(}}{{\rm{h}}^{\rm{2}}}{\rm{ - ab}})} }}{{(a + b)}}$,
We simplify it further to
$ = \dfrac{{\sqrt {(4{h^2} - 4ab)} }}{{(a + b)}}$
so, after putting values in the formula we get,
$ = \dfrac{{\sqrt {(3{a^2} + 10ab + 3{b^2} - 4ab)} }}{{(a + b)}}$
subtracting similar variables
$ = \dfrac{{\sqrt {\left( {3{a^2} + 6ab + 3{b^2}} \right)} }}{{(a + b)}}$
Taking 3 as common
$ = \dfrac{{\sqrt 3 \left( {{a^2} + 2ab + {b^2}} \right)}}{{(a + b)}}$
Replacing the terms inside the bracket using the formula ${a^2} + {b^2} = {(a + b)^2} - 2ab$
$ = \dfrac{{\sqrt {3{{(a + b)}^2}} }}{{(a + b)}}$
Taking squared terms out of under root
$ = \dfrac{{\sqrt 3 (a + b)}}{{(a + b)}}$
Simplifying further and we get
$ = \sqrt 3 $
$ \Rightarrow \theta = {\tan ^{ - 1}}(\sqrt 3 )$
$ = {60^{\circ}}$

Option ‘C’ is correct

Note: We must not neglect the signs and the correct formula should be used to get a valid answer. For inverse notation, it should look like ${\tan ^{ - 1}}x = y$ where the function is $\tan y = x$ . If we write like ${\left( {\tan x} \right)^{ - 1}} = y$ then it is the wrong procedure.