<s>[INST] How to solve $\frac{1}{2\sin50^\circ}+2\sin10^\circ$ I want to solve the expression $\frac{1}{2\sin50^\circ}+2\sin10^\circ$ to get a much simpler and neater result. I have tried to manipulate this expression such as using sum/difference formulas, but it didn't help (and made the expression even more messy).
Here is what I did:
\begin{align}
\frac{1}{2\sin50^\circ}+2\sin10^\circ&=\frac{1}{2\sin(60-10)^\circ}+2\sin10^\circ \\
&= \frac{1}{2\left(\frac{\sqrt{3}}{2}\cdot\cos10^\circ-\frac{1}{2}\cdot\sin10^\circ\right)}+2\sin10^\circ \\
&= \frac{1}{\sqrt{3}\cdot\cos10^\circ-\sin10^\circ}+2\sin10^\circ \\
&= \frac{\sqrt{3}\cdot\cos10^\circ+\sin10^\circ}{\left(\sqrt{3}\cdot\cos10^\circ-\sin10^\circ\right)\left(\sqrt{3}\cdot\cos10^\circ+\sin10^\circ\right)}+2\sin10^\circ \\
&= \frac{\sqrt{3}\cdot\cos10^\circ+\sin10^\circ}{3\cos^210^\circ-\sin^210^\circ}+2\sin10^\circ
\end{align}
But I don't know how to continue at this point. Multiplying in $2\sin10^\circ$ into the fraction is clearly unrealistic as it would result in trignometry of third power. Any help or hint would be appreciated. According to a calculator, the result of this expression should come to a nicely $1$, but I just want to know how to algebraically manipulate this expression to show that it is equal to $1$.
 [/INST] $$\frac{1}{2\sin{50^{\circ}}} + 2 \sin{10^{\circ}}= \frac{1}{2\sin{50^{\circ}}} + 2 \sin{(60^{\circ}-50^{\circ})}=\frac{1}{2\sin{50^{\circ}}} + \sqrt{3} \cos{50^{\circ}} -\sin{50^{\circ}} =$$
$$=\frac{1-2 \sin^2{50^{\circ}}}{2\sin{50^{\circ}}}+\frac{2\sqrt{3} \cos{50^{\circ}}\sin{50^{\circ}}}{2\sin{50^{\circ}}}=\frac{\cos{100^{\circ}}}{2\sin{50^{\circ}}}+\frac{\sqrt{3}\sin{100^{\circ}}}{2\sin{50^{\circ}}} =$$
$$=\frac{\frac{1}{2}\cos{100^{\circ}} + \frac{\sqrt{3}}{2} \sin{100^{\circ}}}{\sin{50^{\circ}}} = \frac{\sin{(30^{\circ}+100^{\circ})}}{\sin{50^{\circ}}}=1$$
</s>