Suppose for definiteness we work with Laplace's equation $\triangledown^{2}u =0$ on the unit disk in $R^{2}$. Assuming things are somewhat smooth, suppose one specified the tangential, instead of the normal, derivative of $u$, i.e. specified $\partial{u} / \partial{\theta}$ on the unit circle. Picking any point $\theta_{0}$ on the circle, one could integrate in $\theta$ to obtain $u(\theta)$ on the circle:

$u(\theta) = \int_{\theta_{0}}^{\theta} \partial{u}/\partial{\theta}(\alpha) d \alpha + u(\theta_{0})$.

So a tangential derivative really specifies a Dirichlet boundary condition. The additive constant in the above integral merely corresponds to a constant solution $u$ in the disk.

Suppose for definiteness we work with Laplace's equation $\triangledown^{2}u =0$ on the unit disk in $R^{2}$. Assuming things are somewhat smooth, suppose one specified the tangential, instead of the normal, derivative of $u$, i.e. specified $\partial{u} / \partial{\theta}$ on the unit circle. Picking any point $\theta_{0}$ on the circle, one could integrate in $\theta$ to obtain $u(\theta)$ on the circle:

$u(\theta) = \int_{\theta_{0}}^{\theta} \partial{u}/\partial{\theta}(\alpha) d \alpha + u(\theta_{0})$.

So a tangential derivative really specifies a Dirichlet boundary condition. The additive constant in the above integral merely corresponds to a constant solution in the disk, i.e. it shifts the solution $u$ by a constant amount.

You can in general freely specify Dirichlet or Neumann conditions, but not both. So take your pick, a tangential or a normal derivative for the boundary condition.