The basic equation of Lagrange is given by, $$\frac{\mathrm d}{\mathrm dt} \frac{\partial L}{\partial \dot{q_j}} - \frac{\partial L}{\partial q_j} = Q_j \tag{1}$$
where $T$ is the kinetic energy, $V$ is the potential energy, $L = T-V$. But at any given instant $$T + V = \text{Constant}\tag{2}$$ (Conservation of energy)
I understand that the criteria for Lagrange application is that the coordinates that define the system are to be,
Independent
Complete
Holonomic
and for a mechanical system $V \neq f(\dot{q_j})$.
Thus, for mechanical systems, the above relation reduces to
$$\frac{\mathrm d}{\mathrm dt} \frac{\partial T}{\partial \dot{q_j}} - \dfrac{\partial T}{\partial q_j} + \frac{\partial V}{q_j} = Q_j . \tag{3}$$
Thus, as the coordinates are to be independent,
The effect of a force in any particular direction (component) will not give a displacement in the other directions, thus the energies in that particular direction can be expressed by isolated equations that are independent of each other (This is what I understood by the coordinates being independent).
Then, by applying
$$V = Constant - T = C - T\tag{4}$$
the above equation would just become
$$\frac{\mathrm d}{\mathrm dt} \frac{\partial T}{\partial \dot{q_j}} - \frac{\partial T}{\partial q_j} + \frac{\partial (C - T)}{\partial q_j} = Q_j \tag{5}$$
and eventually boil down to
$$\frac{\mathrm d}{\mathrm dt} \frac{\partial T}{\partial \dot{q_j}} - 2\frac{\partial T}{\partial q_j} = Q_j \tag{6}$$
as the derivative of a constant is always zero.
I am new to dynamics and so like to know where I am wrong conceptually.