tanA=√3 cotB,求（2+cos2A）(2+cos2B)=?
热心网友
∵tanA=√3 cotB ∴tanA/cotB＝tgA*tgB＝√3（2+cos2A）(2+cos2B)＝[2+(1-tgA^2)/(1+tgA^2)]*[2+(1-tgB^2)/(1+tgB^2)] 万能公式＝[(3+tgA^2)/(1+tgA^2)]*[(3+tgB^2)/(1+tgB^2)]＝(9+3tgA^2+3tgB^2+tgA^2*tgB^2)/(1+tgA^2+tgB^2+tgA^2*tgB^2)＝(9+3tgA^2+3tgB^2+3)/(1+tgA^2+tgB^2+3) ∵tgA*tgB＝√3＝(12+3tgA^2+3tgB^2)/(4+tgA^2+tgB^2)＝[3×(4+tgA^2+tgB^2)]/(4+tgA^2+tgB^2)＝3