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.

\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 $
