If $x\cot x=a_0+a_2x^2+a_4x^4+\cdots$, then $\frac{a_{2n}}{1!}-\frac{a_{2n-1}}{3!}+\cdots+\frac{(-1)^na_0}{(2n+1)!}=\frac{(-1)^n}{(2n)!}$

Question

If $x\cot x=a_0+a_2x^2+a_4x^4+\cdots$, then show that $$ \dfrac{a_{2n}}{1!}-\dfrac{a_{2n-2}}{3!}+\dfrac{a_{2n-4}}{5!}-\cdots+\dfrac{(-1)^na_{0}}{(2n+1)!}=\dfrac{(-1)^n}{(2n)!} $$ This problem is from the book "Higher Trigonometry by B.C DAS & N. MUKHERJEE". Unfortunately, there is no solution manual available on the internet so I have to post here.