A Recurrence type relation with an Integral.

Note that $\begin{aligned} \lambda_{m+2} - \lambda_m & = \; \int_0^\pi \frac{\cos(mx) - \cos((m+2)x)}{1-\cos x}\sin x\,dx \; = \; 2\int_0^\pi \frac{\sin((m+1)x)}{1-\cos x}\,\sin^2x\,dx \\ & = \; 2\int_0^\pi \sin((m+1)x)(1 + \cos x)\,dx \; = \; \int_0^\pi \big(2\sin((m+1)x) + \sin((m+2)x) + \sin(mx)\big)\,dx \\ & = \; \Big[-\tfrac{2}{m+1}\cos((m+1)x) - \tfrac{1}{m+2}\cos((m+2)x) - \tfrac{1}{m}\cos mx\Big]_0^\pi \\ & = \; \left\{ \begin{array}{lll} \tfrac{4}{m+2} & \hspace{1cm} & m \; \mbox{even} \\ \tfrac{2}{m+2} + \tfrac{2}{m} & & m\;\mbox{odd} \end{array}\right. \end{aligned}$ Similarly $\begin{aligned} \lambda_{m+2} - \lambda_{m+1} & = \; \int_0^\pi \frac{\cos((m+1)x) - \cos((m+2)x)}{1 - \cos x}\sin x\,dx \; = \; 2\int_0^\pi \frac{\sin((m+\frac32)x)\sin\frac12x}{1 - \cos x}\sin x\,dx \\ & = \; 2\int_0^\pi \sin((m+\tfrac32)x)\cos\tfrac12x\,dx \; = \; \int_0^\pi \big(\sin((m+1)x) + \sin((m+2)x)\big)\,dx \\ & = \; \Big[-\tfrac{1}{m+1}\cos((m+1)x) - \tfrac{1}{m+2}\cos((m+2)x)\Big]_0^\pi \\ & = \; \left\{ \begin{array}{lll} \tfrac{2}{m+1} & \hspace{1cm} & m \; \mbox{even} \\ \tfrac{2}{m+2} & & m\;\mbox{odd} \end{array}\right. \end{aligned}$ and it is now easy to see that $(m+2)\lambda_{m+2} - 2\lambda_{m+1} - m\lambda_m \; = \; m(\lambda_{m+2} - \lambda_m) - 2(\lambda_{m+1} - \lambda_m) \; = \; \boxed{4}$ for all positive integers $m$ .