Twisted Trig!

$8\cos^{2}10^{\circ}-\dfrac{1}{\sin 10^{\circ}}$

$=4(\cos 20^{\circ}+1)-\dfrac{1}{\sin 10^{\circ}}$

$=\dfrac{4\sin 10^{\circ}\cos 20^{\circ}+4\sin 10^{\circ}-1}{\sin 10^{\circ}}$

$=\dfrac{2(\sin 30^{\circ}-\sin 10^{\circ})+4\sin 10^{\circ}-1}{\sin 10^{\circ}}$

$=\dfrac{-2\sin 10^{\circ}+4\sin 10^{\circ}}{\sin 10^{\circ}} =2$

Firstly the trigonometric identities

$2\cos^2 x=1+\cos 2x$

$1+2\cos 2x=\frac{\sin 3x}{\sin x}$

now

$4(1+\cos 20^{\circ})-\frac{1}{\sin 10^{\circ}}$

$= 4+2(2\cos 20^{\circ})-\frac{1}{\sin 10^{\circ}}$

$=4+2(\frac{\sin 30^{\circ}}{\sin 10^{\circ}}-1)-\frac{1}{\sin 10^{\circ}}$

$=4-2+\frac{1}{\sin 10^{\circ}}-\frac{1}{\sin 10^{\circ}}$

$=\large{\boxed{2}}$