How to derive directly from Hamilton equations to the Euler-Lagrange equations?

Question

We have the function (Lagrangian):$$L(\theta, \phi, \dot{\theta}, \dot\phi)=\frac{mb^2}{2}(\dot{\theta}^2+\sin^2(\theta)\dot{\phi}^2)+mbg\cos(\theta)$$ And I have found the Hamilton function $H$ and their equations to: $$H=T+V=\frac{1}{2}mb^2 \dot{\theta}^2+\frac{1}{2}mg^2sin^2(\theta)\dot{\phi}^2+(-mbg cos(\theta))=\frac{p_{\theta}^2}{2mb^2}+\frac{p_{\phi}^2}{2mb^2 sin^2(\theta)}-mgb cos(\theta)$$ $$\dot{p_{\phi}}=-\frac{\partial H}{\partial \phi} = 0$$ $$\dot{p_{\theta}}=-\frac{\partial H}{\partial \theta} =- (-\frac{p_{\phi}^2 2cos(\theta)}{2mb^2 sin^3(\theta)}-(mgb(-sin(\theta))))= \frac{p_{\phi}^2 cos(\theta)}{mb^2 sin^3(\theta)}-mgbsin(\theta)$$ $$\dot{\theta}=\frac{\partial H}{\partial p_{\theta}} =\frac{2 p_{\theta}}{2mb^2}=\frac{p_{\theta}}{mb^2}$$ $$\dot{\phi}=\frac{\partial H}{\partial p_{\phi}} =\frac{2p_{\phi}}{2mb^2 sin^2(\theta)}=\frac{p_{\phi}}{mb^2 sin^2(\theta)}$$ Now I have to show how to derive directly the Euler-Lagrange equations from these and show that H itself is a constant of motion. Can anyone help me how to do that? Eventually see https://en.wikipedia.org/wiki/Spherical_pendulum?