The correct answer is $\boxed{\theta = {\tan ^{ – 1}}\left( {0.25} \right)}$.
The horizontal range of a projectile is the horizontal distance traveled by the projectile before it hits the ground. The maximum height of a projectile is the highest point reached by the projectile.
The angle of projection is the angle between the initial velocity vector of the projectile and the horizontal.
The formula for the horizontal range of a projectile is $R = \frac{v^2 \sin 2 \theta}{g}$, where $v$ is the initial velocity of the projectile, $\theta$ is the angle of projection, and $g$ is the acceleration due to gravity.
The formula for the maximum height of a projectile is $H = \frac{v^2 \sin^2 \theta}{2g}$.
For a projectile, the horizontal range and maximum height are equal when $\sin 2 \theta = \frac{1}{\sqrt{2}}$. This can be solved for $\theta$ to get $\theta = {\tan ^{ – 1}}\left( {0.25} \right)$.
Option A is incorrect because 45 degrees is not the angle of projection at which the horizontal range and maximum height of a projectile are equal.
Option B is incorrect because $\tan ^{ – 1}(0.25)$ is not an angle.
Option C is incorrect because ${\tan ^{ – 1}}4$ is not an angle.
Option D is incorrect because 60 degrees is not the angle of projection at which the horizontal range and maximum height of a projectile are equal.