How do these two equations below hold?

Question

$$-{2l\over m\pi}cos (m\pi) + {2l\over {m^2\pi ^2}} sin (m\pi)=(-1)^{m+1}{2l\over m\pi}, m\in \mathbb N\cup\{0\}$$

$${2l\over m\pi}sin (m\pi) + {2l\over {m^2\pi ^2}} (cos (m\pi)-1)={2l\over m^2\pi ^2}[(-1)^m-1],m\in \mathbb N$$

Answer 1

not that $$\sin(m\pi)=0$$ and $$-\frac{2l}{m\pi}(-1)^m=(-1)^{m+1}\frac{2l}{m\pi}$$