I am trying to deduce the recurrence relation for the modified Bessel function of the second kind $\mathcal{K}_\nu(x)$ (the answer is shown here, page 20). I am clearly making a mistake but I haven't found it after trying many times.
My starting point is the definition $$ \mathcal{K}_\nu(x)=\frac{\pi}{2}\frac{I_{-\nu}-I_{\nu}}{\sin \nu\pi } \tag{1}$$ where $I_{\nu}(x)$ is the modified Bessel function (of first kind) together recurrence relations for $I_{\nu}(x)$: $$ I_{\nu}(x)=I_{\nu\pm1}'(x)\pm \frac{\nu\pm 1}{x}I_{\nu\pm 1}(x) \tag{2}$$ where $I_\nu'(x)=\frac{d}{dx}I_\nu(x)$.
I then try to compute $\mathcal{K}_\nu(x)$: $$ \mathcal{K}_{\nu+1}=\frac{\pi}{2}\frac{I_{-\nu-1}-I_{\nu+1}}{\sin (\nu+1)\pi }=-\frac{\pi}{2}\frac{I_{-\nu-1}-I_{\nu+1}}{\sin \nu\pi }$$ where we use $\sin (a+\pi)=-\sin (a)$. Then, using $(2)$ we get: $$ \frac{2}{\pi}\mathcal{K}_{\nu+1} =\frac{-1}{\sin \nu\pi}\left(I_{-\nu}'(x)+\frac{\nu}{x}I_{-\nu}(x)-\left[ I_{\nu}'(x)-\frac{\nu}{x}I_{\nu}(x)\right]\right) \\ = -\frac{I_{-\nu}'(x)-I_{\nu}'(x)}{\sin \nu\pi}\tag{3} -\frac{1}{\sin\nu\pi}\frac{\nu}{x}\left(I_{-\nu}(v)+I_{\nu}(x)\right)$$ I identify the first term (of the RHS) as $-\frac{2}{\pi}\mathcal{K}_{\nu}'(x)$ which gives me $$\mathcal{K}_{\nu+1}=-\mathcal{K}_{\nu}'(x)+\frac{\pi}{2\sin \nu\pi}\frac{\nu}{x}(I_{-\nu}(v)+I_{\nu}(x))$$ I expect the last term to be similar to $\mathcal{K}_\nu(x)$ but clearly am making a sign mistake somewhere. Where is this mistake?