The posted diagram is misleading. I thought that the straight members of the system are more massive than the mass m at the end of the pivoted h member.

But the equation that you want to prove implies that the masses of the straight members be ignored. So let us consider the pivoted h member as a sling only.

Let us imagine the mass m as a bob whose mass is m.

The rotational speed of the axle induces centripetal forces to the members of the system proportional to the distance of the members from the axle, and those inward forces are perpendicular to the axle. Since the axle is vertical, then the centripetal forces are horizontal.

On the bob, the centripetal force is

Fx = m*(v^2)/r ------------(i)

where

v = tangential velocity of the bob

r = horizontal distance of the bob from the axle

v = w*r

So,

(v^2)/r = (w*r)^2 / r = (w^2)*r

r = a +h*sin(theta)

Hence,

Fx = m[(w^2)*(a +h*sin(theta)] --------(ii)

On the bob also, the gravity exerts a vertical force

Fy = weight = m*g -------(iii)

Now cut the h string at the pivot. Supply necessary forces to the string at that cut point to keep the severed members in equilibrium. Meaning, draw a free-body diagram of the severed portion of the system.

To ignore whatever are those supplied forces at the cut point, we take moments at that cut point.

The FBD, (free-body diagram), shows two forces acting on the bob:

>>>Fx, which is counteracting the centripetal force Fx, is pointing horizontally away from the axle.

>>>Fy, the vertical force pointing vertically downward because of the weight of the bob.

Summation of moments at the cut point, at the pivot, is zero.

(Fx)*(h*cos(theta)) -(Fy)*(h*sin(theta)) = 0

Substitutions,

[m(w^2)(a +h*sin(theta)](h*cos(theta) -[mg](h*sin(theta)) = 0

(mh)(w^2)cos(theta)[a +h*sin(theta)] = (mh)(g)sin(theta)

The (mh) cancels out, then divide both sides by cos(theta)

(w^2)[a +h*sin(theta)] = (g)sin(theta) / cos(theta)

[a +h*sin(theta)](w^2) = (g)tan(theta) --------------proven.