Prove that:
$$
\frac{\sin(120^{\circ})}{1+\sin(90^{\circ}+\mathbf{a})}-\frac{\sin(240^{\circ})}{1-\cos(\mathbf{-a})}=\frac{\sqrt3}{\sin^2(\mathbf{a})}
$$
I am having difficulty converting the LHS into the RHS. I have tried applying trigonometric identities, rationalising the denominator and trying to equate it through working on both sides. However, I am still unable to find the method of solving it. Any help would be appreciated.
Best Answer
We have that $\sin(90^{\circ}+\mathbf{a})=\cos(\mathbf{-a})=\cos(\mathbf{a})$. Moreover $$\sin^2(\mathbf{a})=1-\cos^2(\mathbf{a})=(1+\cos(\mathbf{a}))(1-\cos(\mathbf{a})).$$ Therefore $$\frac{\sin(120^{\circ})}{1+\sin(90^{\circ}+\mathbf{a})}-\frac{\sin(240^{\circ})}{1-\cos(\mathbf{-a})}= \frac{\sin(120^{\circ})(1-\cos(\mathbf{a}))-\sin(240^{\circ})(1+\cos(\mathbf{a}))}{\sin^2(\mathbf{a})}\\=\frac{\sqrt3}{\sin^2(\mathbf{a})}$$ where in the last step we used $\sin(120^{\circ})=-\sin(240^{\circ})=\sqrt{3}/2$