How to show that,
$$ I_{\nu}(x) K^{'}_{\nu}(x) - I^{'}_{\nu}(x) K_{\nu}(x) = -\frac{1}{x} $$
where, $I_{\nu}$ and $K_{\nu}$ are modified Bessel functions of first and second kind, respectively.
Edit: I did some work and brought the left hand side of equation above to
$$ \frac{\pi}{x\sin\nu\pi} \left[ \sum_{s=0}^\infty \sum_{m=0}^\infty \frac{s+m}{s!m!(s+\nu)!(m-\nu)!} \left(\frac{x}{2}\right)^{2s+2m} \right] $$
Now the summation inside parenthesis should be equal to $-\frac{\sin\nu\pi}{\pi}$ in order to get the desired result. But I don't know how to derive it. Any help?
Edit2: I tried another way but got stuck again.
\begin{align*} I_\nu(x)K^{'}_\nu(x) - I^{'}_\nu(x)K_\nu(x) &= \frac{\pi}{2} \frac{I_\nu I^{'}_{-\nu}-I_\nu I^{'}_{\nu}}{\sin\nu\pi} - \frac{\pi}{2} \frac{I^{'}_\nu I_{-\nu}-I^{'}_\nu I_{\nu}}{\sin\nu\pi} \\ &= \frac{\pi}{2\sin\nu\pi} \left[ I_\nu I^{'}_{-\nu}-I_\nu I^{'}_{\nu} - I^{'}_\nu I_{-\nu}+I^{'}_\nu I_{\nu} \right] \\ &= \frac{\pi}{2\sin\nu\pi} \left[ I_\nu I^{'}_{-\nu} - I^{'}_\nu I_{-\nu} \right] \\ &= \frac{\pi}{2\sin\nu\pi} \left[ I_\nu (I^{'}_{-\nu} - I^{'}_\nu) \right] \end{align*}
The modified Bessel functions of order $\nu$, labeled as $I_\nu,K_\nu$, satisfy modified Bessel differential equation $x^2 y''+xy'-(x^2-\nu^2)y=0$. This allows us to simplify the derivative of the Wronskian $W(x)=I_{\nu} K'_{\nu} - I'_{\nu} K_{\nu}$:
\begin{align} W'(x) &=(I'_\nu K'_\nu+I_\nu K''_\nu) -(I''_\nu K_\nu+I'_\nu K'_\nu)\\ &=I_\nu K''_\nu-I''_\nu K_\nu\\ &=I_\nu \cdot \frac{1}{x^2}\left[(x^2-\nu^2)K_\nu-x K'_\nu\right] -\frac{1}{x^2}\left[(x^2-\nu^2)I_\nu-x I'_\nu\right]\cdot K_\nu\\ &=\frac{1}{x}\left(I_\nu'K_\nu-I_\nu K'_\nu\right).\\ \end{align} Hence $W'(x)=-W(x)/x,$ which has the general solution $W(x)=A/x$. All that remains is show $A=-1,$ which follows from the asymptotic behavior at $x=\infty$.