Sum of series of coefficients of zeta in even integers

Question

Prove that the $\sum (2n+1) b_{2n}=\frac{1-\sin2}{1-\cos 2}$, where $b_{2n}=\pi^{-2n} \zeta(2n)$

I found in book that $b_{2m}-\frac{1}{3!}b_{2m-2}+ \frac{1}{5!}b_{2m-4}+...+(-1)^{m-1}\frac{1}{(2m-1)!}b_{2}=(-1)^{m-1}\frac{m}{(2m+1)!}$
and proved by myself that $\sum b_{2n}=\frac{\sin 1 -\cos 1}{2 \sin 1}$

Answer 1

Since $$\sum_{n\geq2}x^{n}\zeta\left(n\right)=-\gamma x-x\psi\left(1-x\right),\,-1\leq x<1$$ where $\gamma$ is the Euler-Mascheroni constant and $\psi(x)$ is the digamma function we have, by the multisection formula and the reflection formula, that $$\sum_{n\geq1}x^{2n}\zeta\left(2n\right)=\frac{1}{2}\left[1-\pi x\cot\left(\pi x\right)\right].$$ Hence $$\sum_{n\geq1}\left(2n+1\right)\pi^{-2n}\zeta\left(2n\right)=\frac{d}{dx}\left(\frac{x}{2}\left[1-\pi x\cot\left(\pi x\right)\right]\right)_{x=1/\pi}=\color{red}{\frac{1}{2}\left(1+\frac{1}{\sin^{2}\left(1\right)}-\frac{2\cos\left(1\right)}{\sin\left(1\right)}\right)}.$$