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Solution

For this expression, $\dfrac{1}{\sin 10^\circ} - \dfrac{\sqrt{3}}{\cos 10^\circ} = \dfrac{\cos 10^\circ - \sqrt{3}\sin 10^\circ}{\sin 10^\circ \cos 10^\circ}.$ Now, $\cos 10^\circ - \sqrt{3}\sin 10^\circ = 2\cos(60^\circ + 10^\circ) = 2\cos 70^\circ.$ Also, $\sin 10^\circ \cos 10^\circ = \dfrac{1}{2}\sin 20^\circ.$ Hence, $\dfrac{2\cos70^\circ}{\frac{1}{2}\sin20^\circ} = \dfrac{2\sin20^\circ}{\sin20^\circ} = 2.$