Consider a SDE system for Brownian motion on the sphere in spherical coordinates, in particular $$d \theta_t = \frac12 \cot \theta_t dt + dB_t^1$$ $$d \phi_t =\frac{1}{\sin \theta_t} dB_t^2$$ where $B=(B^1,B^2)$ is a standard planar BM.
If we let $A=(\theta, \phi)^T$ we can write this in vector form $$dA_t=\mu(A)dt+\sigma(A) dB_t$$ where $\mu =(\frac12 \cot \theta, 0)^T$ and $$\sigma = \begin{pmatrix} 1 & 0 \\ 0 & \frac{1}{\sin \theta} \\ \end{pmatrix}$$
To convert to Stratonovich, we need to compute $$2c_i=\sum_k \sum_j \sigma_{kj} \frac{\partial \sigma_{ij}}{\partial x_k}$$ where $(x_1, x_2)=(\theta, \phi)$ for notational convenience. Since the off diagonals are zero and its a 2 dimensional case we simplify this as $$2c_i = \frac{\partial \sigma_{i1}}{\partial \theta} +\frac{1}{\sin \theta}\frac{\partial \sigma_{i2}}{\partial \phi}$$ Substituting the values of $\sigma$ and computing the derivatives we see that $c$ is the zero vector identically. Therefore the Stratonovich SDE is identical (same drift, same diffusion coefficients) to the Itô SDE.
My Question Are these computations and is this conclusion correct?