The value of $$\dfrac{\sqrt{3}  \operatorname{cosec} 20^{\circ}-\sec20^{\circ}}{\cos20^{\circ}\cos40^{\circ}\cos60^{\circ}\cos80^{\circ}}$$ is equal to:

We know that - 
$$\sin(2A)=2\sin A\cos A$$
$$\sin(A-B)=\sin A\cos B-\cos A\sin B$$
$$\cos60^o=\dfrac{1}{2}$$

Thus, the given equation will be - 

$$=\dfrac{\sqrt{3} \cosec20^{\circ}-\sec20^{\circ}}{\cos20^{\circ}\cos40^{\circ}\cos60^{\circ}\cos80^{\circ}}$$

$$=\dfrac{\frac{\sqrt{3}}{\sin20^o}-\frac{1}{\cos20^o}}{\cos20^o\cos40^o\cos60^o\cos80^o}$$

$$=\dfrac{\frac{\sqrt{3}\cos20^o-\sin20^o}{\sin20^o\cos20^o}}{\cos20^o\cos40^o\cos60^o\cos80^o}$$

$$=\dfrac{\sqrt{3}\cos20^o-\sin20^o}{\left(\sin20^o\cos20^o\right)\cos20^o\cos40^o\cos60^o\cos80^o}$$

$$=\dfrac{\sqrt{3}\cos20^o-\sin20^o}{\left(\frac{\sin40^o}{2}\right)\cos20^o\cos40^o\cos60^o\cos80^o}$$

$$=\dfrac{2\left(\sqrt{3}\cos20^o-\sin20^o\right)}{\left(\sin40^o\cos40^o\right)\cos20^o\cos60^o\cos80^o}$$

$$=\dfrac{2\left(\sqrt{3}\cos20^o-\sin20^o\right)}{\left(\frac{\sin80^o}{2}\right)\cos20^o\cos60^o\cos80^o}$$

$$=\dfrac{2\times2\left(\sqrt{3}\cos20^o-\sin20^o\right)}{\sin80^o\cos80^o\cos20^o\times\frac{1}{2}}$$

$$=\dfrac{2\times2\left(\sqrt{3}\cos20^o-\sin20^o\right)}{\frac{\sin160^o}{2}\times\cos20^o\times\frac{1}{2}}$$

$$=\dfrac{2\times2\times2\times2\left(\sqrt{3}\cos20^o-\sin20^o\right)}{\sin160^o\times\cos20^o}$$

$$=\dfrac{2\times2\times2\times2\left(\sqrt{3}\cos20^o-\sin20^o\right)}{\sin\left(180-20\right)^o\times\cos20^o}$$

$$=\dfrac{2\times2\times2\times2\left[2\times\left(\frac{\sqrt{3}}{2}\cos20^o-\frac{1}{2}\sin20^o\right)\right]}{\sin20^o\times\cos20^o}$$

$$=\dfrac{2\times2\times2\times2\times2\left[\sin60^o\cos20^o-\cos60^o\sin20^o\right]}{\frac{\sin40^o}{2}}$$

$$=\dfrac{2\times2\times2\times2\times2\times2\left[\sin40^o\right]}{\sin40^o}$$

$$=64$$