$T_n(x)$ is the Chebyshev polynomials of the first kind. I was wondering how to show the following recurrence relationship: $$ (1-x^2)T'_k + kxT_k -kT_{k-1} = 0$$
This recurrence form also shows up as equation (2) in this paper: https://www.mathsjournal.com/pdf/2016/vol1issue1/PartA/1-1-18-475.pdf
But it doesn't seem to provide the proof.
Edit: Is it possible to prove it using only the three-term recurrent relation: $T_{k+1} = 2xT_k - T_{k-1}$?
Thanks in advance.
Plugging in $x=\cos t$, $$\begin{split} (1-x^2)T'_k(x) + kxT_k(x) -kT_{k-1}(x) &=(1-\cos^2 t)T^\prime_k(\cos t) + k\cos(t) T_k(\cos t) -kT_{k-1}(\cos t) \\ &= \sin^2(t) \frac{k\sin(kt)}{\sin t}+k\cos(t)\cos(kt)-k\cos((k-1)t)\\ &= k\sin(t)\sin(kt)+k\cos(t)\cos(kt)\\ &-k\cos(kt)\cos t -k\sin(kt)\sin t \\ &=0 \end{split}$$ where we have used the fact that $T_n(\cos t)=\cos(nt)$ and by differentiation, $$(\sin t) T^\prime_k(\cos t)=n\sin(nt)$$ We have thus proven that for all $x\in[-1, 1]$, $$(1-x^2)T'_k(x) + kxT_k(x) -kT_{k-1}(x) = 0$$ This implies that this identity must be true as a polynomial equality.