tan80º+tan40º-√3tan80ºtan40º的值是

= sin80/cos80 sin130（1+√3tan80） = 2·(sin130/sin80)·[(1/2)·sin80 + (√3/2)·cos80] = 2·(sin130/sin80)·[sin(80 + 60)] = 2·(sin130/sin80)·sin140 = 2·(sin130/sin80)·sin40 = 2·sin130·sin40/(2sin40·cos40) = sin130/cos40 =...

(1-tan40°tan80°)/(cot50°+cot10°）=(cos40cos80-sin40sin80)/cos40cos80/((cos50sin10+sin50cos10)/sin50sin10)=cos(40+80)/sin(50+10)=-cos60/sin60 =-√3/3

用这个公式化简tan(a+b) =(tana + tanb)/(1-tana·tanb)原式=tan40°tan80°-根号3·tan120°·（1-tan40°tan80°）=tan40°tan80°+3（1-tan40°tan80°）=3-2·tan40°tan80°