To prove $\frac{1}{\sin 10^\circ}-\frac{\sqrt 3}{\cos 10^\circ}=4$ To prove:
$$\frac{1}{\sin 10^\circ}-\frac{\sqrt 3}{\cos 10^\circ}=4$$
I tried taking lcm but could not get to anything.
 A: \begin{align*}
\frac {1}{\sin 10^\circ} - \frac {\sqrt 3}{\cos 10^\circ} &= \frac {\cos 10^\circ - \sqrt 3\sin 10^\circ}{\sin 10^\circ \cos 10^\circ} \\
&= 2\left(\frac {\frac 12\cos 10^\circ - \frac {\sqrt 3}{2}\sin 10^\circ}{\sin 10^\circ \cos 10^\circ}\right) \\
&= 2\left(\frac {\sin 30^\circ\cos 10^\circ - \cos 30^\circ\sin 10^\circ}{\sin 10^\circ \cos 10^\circ}\right) \\
(a) \ldots &= 2\left(\frac {\sin (30^\circ-10^\circ)}{\sin 10^\circ \cos 10^\circ}\right) \\
&= 2\left(\frac {\sin 20^\circ}{\sin 10^\circ \cos 10^\circ}\right) \\
(b) \ldots &= 2\left(\frac {2\sin 10^\circ \cos 10^\circ}{\sin 10^\circ \cos 10^\circ}\right) \\
&= 4
\end{align*}
$(a) \ldots \rightarrow \sin (X-Y)=\sin X \cos Y - \sin Y \cos X $
$(b) \ldots \rightarrow \sin (2X)=2\sin X \cos X $
A: Set
$$
x=\dfrac{1}{\sin10^\circ}-\dfrac{\sqrt{3}}{\cos10^\circ},\quad y=\sin10^\circ\cos10^\circ=\dfrac12\sin20^\circ \ne 0.
$$
We have
\begin{eqnarray}
\dfrac12xy&=&\dfrac12\cos10^\circ-\dfrac{\sqrt{3}}{2}\sin10^\circ=\sin30^\circ\cos10^\circ-\cos30^\circ\sin10^\circ=\sin(30^\circ-10^\circ)=\sin20^\circ,
\end{eqnarray}
i.e.
$$
\dfrac12xy=2y.
$$
Since $y\ne 0$, it follows that
$$
x=4.
$$
A: $$ \frac{\sin 20}{\sin20}=1$$
$$\frac {\sin(30-10)}{\frac12\sin 10 \cos 10}=1 $$
 and hence after expanding $\sin(30-10)$ we can have the result.
