Lets recall the following identities:
- $\sin\left( 360 - x \right) = - sin\left(x\right)$
- $\cos\left(-x\right) = \cos\left(x\right)$
- $\sin\left( 90 + x \right) = \cos\left( x \right)$
- ${\sin\left( x \right)}^2 + {\cos\left( x \right)}^2 = 1$
Now since we have these identities, we can prove as follows:
$$\frac{\sin(120^{\circ})}{1+\sin(90^{\circ}+\mathbf{a})}-\frac{\sin(240^{\circ})}{1-\cos(\mathbf{-a})}=\frac{\sin(120^{\circ})}{1+\sin(90^{\circ}+\mathbf{a})}+\frac{\sin(120^{\circ})}{1-\cos(\mathbf{-a})},$$
where the equality is the first identity.
Using the second and third identities we yield,
$$\frac{\sin(120^{\circ})}{1+\sin(90^{\circ}+\mathbf{a})}+\frac{\sin(120^{\circ})}{1-\cos(\mathbf{-a})} = \frac{\sin(120^{\circ})}{1+\cos(a)}+\frac{\sin(120^{\circ})}{1-\cos(\mathbf{a})} = \frac{2\cdot \sin(120^{\circ})}{1-{\cos(a)}^2}$$
By the last identity, we obtain
$$\frac{2\cdot \sin(120^{\circ})}{1-{\cos(a)}^2} = \frac{2\cdot \sin(120^{\circ})}{{\sin(a)}^2}$$
Since $\sin(120^{\circ}) = \frac{\sqrt{3}}{2}$, then combining all the previous equalities yields
$$\frac{\sin(120^{\circ})}{1+\sin(90^{\circ}+\mathbf{a})}-\frac{\sin(240^{\circ})}{1-\cos(\mathbf{-a})}=\frac{\sqrt3}{\sin^2(\mathbf{a})}$$