Question

prove that tan20 .tan40.tan80=tan60.

Answer 1

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

Answer 2

tan 40 * tan 80* tan 20=(sin 40 *sin 80) * sin 20 /(cos 40 *cos 80 ) cos 20

now, multiplying numerator and denominator by 2,

=(2*sin 40*sin 80) *sin 20 /(2 cos 40*cos 80) cos 20]

=(cos 40- cos 120)*sin 20/( (cos 120+ cos 40) cos 20

=(2 cos 40 +1) sin 20 /(2 cos 40-1) cos 20

=(2 cos 40*sin 20 +sin 20) /(2 cos 40 cos 20 -cos 20)

=(sin 60 -sin 20 +sin 20)/ (cos 60+sin 20-sin 20)

= sin 60/cos 60

=tan 60