Question

Prove that tan20 tan40 tan60 tan80 = 3

Answer 1

tan20°tan40°tan60°tan80°
={(sin20°sin40°sin80°)/(cos20°cos40°cos80°)}×√3
=√3×{(2sin20°sin40°sin80°)/(2cos20°cos40°cos80°)}
=√3×[{cos(20°-40°)-cos(20°+40°)}sin80°/{cos(20°-40°)+cos(20°+40°)}cos80°]
=√3×[{(cos20°-cos60°)sin80°}/(cos20°+cos60°)cos80°}]
=√3×[{cos20°sin80°-(1/2)sin80°}/{cos20°cos80°+(1/2)cos80°}] [∵, cos60°=1/2]
=√3×[{2cos20°sin80°-2×(1/2)×sin80°}/{2cos20°cos80°+2×(1/2)×cos80°}]
=√3×[{sin(20°+80°)-sin(20°-80°)-sin80°}/{cos(20°-80°)+cos(20°+80°)+cos80°}]
=√3×[{sin100°-sin(-60°)-sin80°}/{cos60°+cos100°+cos80°}]
=√3×[{sin100°+sin60°-sin80°}/{(1/2)+cos100°+cos80°}]
=√3×[{sin100°-sin80°+√3/2}/{1/2+cos100°+cos80°}]
=√3×[{2cos(100°+80°)/2sin(100°-80°)/2+√3/2}/{1/2+2cos(100°+80°)/2cos(100°-80°)/2}]
=√3×[{2cos90°sin10°+√3/2}/{1/2+2cos90°cos10°}]
=√3×{(√3/2)/(1/2)} [∵, cos90°=0]
=√3×√3/2×2/1
=3 (Proved)

Answer 2

using tan(3x) formula

tan3x= (3tanx-tan$^{3}$x)/(1-3$tan^{2} x$)