string theory - Reduction of Nambu-Goto action to true degrees of freedom

I) First consider the point particle

$$S=m\int\sqrt{-\dot{X}^2}d\tau.$$

If you choose the static gauge $$\tau=X^0$$ and replace it in the action you get

$$=m\int\sqrt{1-\dot{X}^j\dot{X}^j}d\tau.$$

So now, you have an equivalent action with true degrees of freedom only. In fact, you can do the same with the light-cone gauge $$\tau=X^+$$ and obtain

$$S=m\int\sqrt{2\dot{X}^--\dot{X}^I\dot{X}^I}d\tau.$$

II) Is it possible to do the same for the string? I have found no reference doing this. I have been trying replacing the conditions of the light-cone gauge. I think that the answer could be

$$\mathcal{L}\sim \dot{X}^I\dot{X}^I -{X^I}' {X^I}'.\tag{12.81} $$

This is eq.(12.81) in the 2nd edition of Zwiebach's book.

Additional Information: Basically you have to show that the gauge conditions $$n\cdot X=2\beta \alpha'n\cdot p \tau ,\tag{1}$$

$$n\cdot p=\frac{2\pi}{\beta}n\cdot \mathcal{P},\tag{2}$$

(where $\beta=2$ for open string and $\beta=1$ for closed string) imply $$\dot{X^2}= -X'^2\quad\text{and}\quad \dot{X}\cdot X'=0.$$

And with these relations you can easily reduce the action.

For the open string you can show that condition (2) implys $$\dot{X}\cdot X'=0.$$

This can be done using the boundary conditions of the open string.$^1$

But for the closed string I have found no way to show that. In Zwiebach page 180 there is an idea of this for open and closed strings.$^2$

To sum up, my question is if one can deduce the same for the closed string. What would be the procedure?

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$^1$See Sundermayer, Constrained Dynamics page 218, or Hansen-Regge-Teitelboim, Constrained Hamiltonian Systems page 58.

$^2$ But in the case of closed string the full $\sigma=0$ line is constructed by requiring that at each point its tangent be orthogonal to $X'$. So it is like he is imposing $\dot{X}\cdot X'=0$, not deducing it.