The value of $$\operatorname{cosec}10°-\sqrt{3}\sec10°$$ is equal to :

We have, 

$$\operatorname{cosec}10°-\sqrt{3}\sec10°\ =\dfrac{1}{\sin10^{\circ\ }}-\dfrac{\sqrt{\ 3}}{\cos10^{\circ\ }}$$

$$=\dfrac{\cos10^{\circ\ }-\sqrt{\ 3}\sin10^{\circ\ }}{\sin10^{\circ\ }\cos10^{\circ\ }}$$

$$=2\times\ 2\times\ \dfrac{\frac{1}{2}\left(\cos10^{\circ\ }-\sqrt{\ 3}\sin10^{\circ\ }\right)}{2\times\ \sin10^{\circ\ }\cos10^{\circ\ }}$$

$$=4\times\ \dfrac{\frac{1}{2}\cos10^{\circ\ }-\dfrac{\sqrt{\ 3}}{2}\sin10^{\circ\ }}{2\times\sin10^{\circ\ }\cos10^{\circ\ }}$$

$$=4\times\ \dfrac{\sin30^{\circ\ }\cos10^{\circ\ }-\cos30^{\circ\ }\sin10^{\circ\ }}{2\sin10^{\circ\ }\cos10^{\circ\ }}$$

$$=4\times\ \dfrac{\sin\left(30-10\right)^{\circ\ }}{\sin\left(2\times\ 10\right)^{\circ\ }}=4\times\ \dfrac{\sin20^{\circ\ }}{\sin20^{\circ\ }}=4$$

The correct answer is Option A (4).