Question

prove that: tan20.tan30.tan40.tan80=1

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°

Now, tan30°tan60° = $\frac{1}{\sqrt{3} }$*$\sqrt{3}$= 1

Answer 2

Answer:

Step-by-step explanation:

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°

Now, tan30°tan60° = *= 1

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