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Tan20°tan30°tan40° tan80° |
| 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\xa0OR\xa0 | |