I've been told there exists the identity
$$\partial_\theta Y_{\ell,m}(\theta,\phi)=\frac{1}{2}e^{-i\phi}\sqrt{(\ell-m)(\ell+m+1)}Y_{\ell,m+1}-\frac{1}{2}e^{i\phi}\sqrt{(\ell+m)(\ell-m+1)}Y_{\ell,m-1},$$
but can't quite figure out the proof. If I take the derivatives of $Y_{\ell,m}$ and separately $Y_{\ell,-m}$, I pick up complex exponential factors and other things I don't want. Can anyone point me in the right direction? Here is what I have so far: $$\partial_\theta Y_{\ell,m}(\theta,\phi)=e^{-i\phi}\sqrt{(\ell-m)(\ell+m+1)}Y_{\ell,m+1}$$ which is half of it. The second half is the problem. \begin{align}\partial_\theta Y_{\ell,-m}(\theta,\phi)&=e^{-i\phi}\sqrt{(\ell+m)(\ell-m+1)}Y_{\ell,1-m}\nonumber\\ &=e^{-i\phi}\sqrt{(\ell+m)(\ell-m+1)}(-1)^{m-1} Y_{\ell,m-1}^*\\ &=-e^{-i\phi}\sqrt{(\ell+m)(\ell-m+1)}(-1)^{m} Y_{\ell,m-1}^* \end{align}
Perhaps there is another identity I'm unaware of?
First of all, I forgot a piece of the derivative i.e. $$\partial_\theta Y_{\ell,m}(\theta,\phi)=m \cot\theta Y_{\ell,m}(\theta,\phi)+e^{-i\phi}\sqrt{(\ell-m)(\ell+m+1)}Y_{\ell,m+1}(\theta,\phi)$$ Using this result, I simply define $$\partial_\theta Y_{\ell,m}\equiv\frac{1}{2}(\partial_\theta Y_{\ell,+m}+\partial_\theta Y_{\ell,-m}).$$ Now as above, we have $$\partial_\theta Y_{\ell,+m}(\theta,\phi)=m \cot\theta Y_{\ell,m}(\theta,\phi)+e^{-i\phi}\sqrt{(\ell-m)(\ell+m+1)}Y_{\ell,m+1}(\theta,\phi).$$ Using this result, we do the other sign. \begin{align} \partial_\theta Y_{\ell,-m}(\theta,\phi)&=-m \cot\theta Y_{\ell,-m}(\theta,\phi)+e^{-i\phi}\sqrt{(\ell+m)(\ell-m+1)}Y_{\ell,1-m}(\theta,\phi)\nonumber\\ (-1)^m\partial_\theta Y_{\ell,+m}^*(\theta,\phi)&=-(-1)^m m \cot\theta Y_{\ell,+m}^*(\theta,\phi)+e^{-i\phi}\sqrt{(\ell+m)(\ell-m+1)}Y_{\ell,m-1}^*(-1)^{m-1}(\theta,\phi)\nonumber\\ \partial_\theta Y_{\ell,m}^*(\theta,\phi)&=-m \cot\theta Y_{\ell,m}^*(\theta,\phi)-e^{-i\phi}\sqrt{(\ell+m)(\ell-m+1)}Y_{\ell,m-1}^*(\theta,\phi)\nonumber\\ \partial_\theta Y_{\ell,m}(\theta,\phi)&=-m \cot\theta Y_{\ell,m}(\theta,\phi)-e^{i\phi}\sqrt{(\ell+m)(\ell-m+1)}Y_{\ell,m-1}(\theta,\phi)\nonumber\\ \end{align} where in the last step, I took the complex conjugate of both sides. It is then easily seen that \begin{align} \partial_\theta Y_{\ell,m}&\equiv\frac{1}{2}(\partial_\theta Y_{\ell,+m}+\partial_\theta Y_{\ell,-m})\nonumber\\ &=e^{-i\phi}\sqrt{(\ell-m)(\ell+m+1)}Y_{\ell,m+1}(\theta,\phi)-e^{i\phi}\sqrt{(\ell+m)(\ell-m+1)}Y_{\ell,m-1}(\theta,\phi) \end{align} as desired.