Series

tan10°. tan80° =tan 10°.cot10° =1. In such way, tan15°.tan75°=1, tan20°.tan 70°=1, ... ... tan45°.tan 45°=1. So we can say, tan5°.1.1.1.1.1.1.1.1= tan5°

tan(80) = sin(80)/cos(80) = sin(90 - 10)/cos(90 - 10) = cos(10)/sin(10). So tan(10) tan(80) = sin(10) cos(10)/[sin(10)*cos(10)] = 1. All pairings from 10 to 80 give a product of 1 and we are left with tan(5).

$\tan5^{\circ}\cdot \tan10^{\circ} \cdot \tan15^{\circ} \cdots \tan80^{\circ}$

$=\tan5^{\circ}\frac{\sin10^{\circ} \cdot \sin15^{\circ} \cdots \sin80^{\circ}}{\cos10^{\circ} \cdot \cos15^{\circ} \cdots \cos80^{\circ}}$

Because $\sin x = \cos(90^{\circ}-x)$ , we have

$=\tan5^{\circ}\frac{\sin10^{\circ} \cdot \sin15^{\circ} \cdots \sin80^{\circ}}{\sin10^{\circ} \cdot \sin15^{\circ} \cdots \sin80^{\circ}}$

$=\tan5^{\circ}$