# Proving $\frac{\sin(120^{\circ})}{1+\sin(90^{\circ}+\mathbf{a})}-\frac{\sin(240^{\circ})}{1-\cos(\mathbf{-a})}=\frac{\sqrt3}{\sin^2(\mathbf{a})}$

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.

• Hint: Write all the denominators in terms of $\cos a$. (Then factor the denominator on the right-hand side.) – Blue Sep 5 '16 at 10:46

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 $\sin(120^{\circ})=-\sin(240^{\circ})=\sqrt{3}/2$

• I see. The key step I could not pick up on was sin(90∘+a)=cos(−a). I understand now. – Wanyu Tang Sep 5 '16 at 11:06

Use the relations: $\sin\big(90^\circ+a \big)=\cos a$, $\cos a=\cos (-a)$ and $\sin(120^\circ)=-\sin(240^\circ)$. Then the LHS of identity become

$$\frac{\sin(120^\circ)}{1+\cos a}+\frac{\sin(120^\circ)}{1-\cos a}=\frac{2\sin(120^\circ)}{\sin^2 a}=\frac{2\cos(30^\circ)}{\sin^2 a}=\frac{2\frac{\sqrt{3}}{2}}{\sin^2a}=\frac{\sqrt{3}}{\sin^2a}.$$

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 $$\frac{\sin(120^{\circ})}{1+\sin(90^{\circ}+\mathbf{a})}-\frac{\sin(240^{\circ})}{1-\cos(\mathbf{-a})}=\frac{\sqrt3}{\sin^2(\mathbf{a})}$$