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 $1$.
Best Answer
$$\begin{align}\frac1{2\sin50^\circ}+2\sin10^\circ&=\frac{1+4\sin50^\circ\sin10^\circ}{2\sin50^\circ}\\ &=\frac{1+2(\cos40^\circ-\cos60^\circ)}{2\cos40^\circ}\\&=1 \end{align}$$