Solving $E=\frac{1}{\sin10^\circ}-\frac{\sqrt3}{\cos10^\circ}$ $$E=\frac{1}{\sin10^\circ}-\frac{\sqrt3}{\cos10^\circ}$$
I got no idea how to find the solution to this. Can someone put me on the right track?
Thank you very much.
 A: Divide both terms by two and use the fact $\sin(30) = \frac{1}{2}$ and 
$\cos(30) = \frac{\sqrt{3}}{2}$. Then you just need to use the formulas for 
$\sin(a+b)$ and $\sin(a-b)$ to find the solution.
A: We have
\begin{eqnarray*}
E&=&\frac{\cos 10^\circ-\sqrt{3}\sin 10^\circ}{\sin 10^\circ \cos 10^\circ}\\
&=&4\frac{(1/2)\cos 10^\circ-(\sqrt{3}/2)\sin 10^\circ}{2\sin 10^\circ \cos 10^\circ}\\
&=&4\frac{\cos 60^\circ\cos 10^\circ-\sin 60^\circ\sin 10^\circ}{\sin 20^\circ}\\
&=&4\frac{\cos(60^\circ+10^\circ)}{\sin 20^\circ}\\
&=&4\frac{\cos 70^\circ}{\sin 20^\circ}\\
&=&4\frac{\sin 20^\circ}{\sin 20^\circ}\\
&=&4.
\end{eqnarray*}
A: $$\text{Let us check the value of }\frac a{\sin\theta}+\frac b{\cos\theta}$$
$$\frac a{\sin\theta}+\frac b{\cos\theta}=\frac{a\cos\theta+b\sin\theta}{\cos\theta\sin\theta}$$
Putting $a=r\sin\alpha,b=r\cos\alpha$ where $r>0$
Squaring & adding we get $r^2=a^2+b^2\implies r=+\sqrt{a^2+b^2}$
$$\implies \frac a{\sin\theta}+\frac b{\cos\theta}=\frac{2\sqrt{a^2+b^2}(\sin\theta\cos\alpha+\cos\theta\sin\alpha)}{\sin2\theta}$$
$$=2\sqrt{a^2+b^2}\cdot\frac{\sin(\theta+\alpha)}{\sin2\theta}\text{ as }\sin2\theta=2\sin\theta\cos\theta$$
Now, the solution of $P\sin x= Q\sin A $ is general intractable unless  $P=0$ or $Q=0$ or $P=\pm Q\ne0$
Here the coefficients of $\sin(\theta+\alpha),\sin2\theta$ can not be $0$ 
So, either $\sin2\theta=\sin(\theta+\alpha)$ or $\sin2\theta=-\sin(\theta+\alpha)$
$$\begin{array}{|c|c|c|} 
 \hline \text{ Case } & \sin2\theta=\sin(\theta+\alpha) & \sin2\theta=-\sin(\theta+\alpha)=\sin(-\theta-\alpha)  \text{ as }\sin(-x)=-\sin x      \\ 
 \hline \text{General Solution} &  2\theta=n180^\circ+(-1)^n(\theta+\alpha)\text{  where }n\text{ is any integer } & 2\theta=n180^\circ+(-1)^n(-\alpha-\theta)\text{  where }n\text{ is any integer }    \\ 
 \hline n=2m & \alpha=\theta-m360^\circ\equiv\theta\pmod{360^\circ} & \alpha=m360^\circ-3\theta\equiv-3\theta \\
\hline n=2m+1 & \alpha=(2m+1)180^\circ-3\theta\equiv 180^\circ-3\theta & \alpha=\theta-(2m+1)180^\circ\equiv\theta+180^\circ  \\
 \hline  
  \end{array} $$
Here $a=1,b=-\sqrt3$ and $\theta=10^\circ$
Taking $\sin2\theta=\sin(\theta+\alpha), \alpha=\theta=10^\circ$ or $=180^\circ-3\theta=150^\circ$
$\implies r=+\sqrt{a^2+b^2}=2$ and $\cos \alpha=\frac br=-\frac{\sqrt3}2$ and $\sin\alpha=\frac ar=\frac12\implies \alpha$ lies in the 2nd Quadrant, $\implies \alpha=150^\circ$
$$\text{So,} \frac{1}{\sin10^\circ}-\frac{\sqrt3}{\cos10^\circ}=2\sqrt{1^2+(-\sqrt3)^2}\cdot\frac{\sin(10^\circ+150^\circ)}{\sin(2\cdot10^\circ)}=2\cdot2=4$$
