find the angle of projection at which the horizontal range and maximum height of a projectile are

Let θ be the angle of projection for which horizontal range and maximum height of a projectile are equal.Now,

H= u² sin² θ / 2g

R = u² sin2θ / g

So, if H=R

then , u² sin² θ / 2g = u² sin2θ / g

or, sin² θ / 2 = sin2θ

or, sin² θ  =  2 (2 sinθ  cosθ )

or tan θ = 4

or, θ  = tan ^-1 ( 4)