tan1°tan2°+tan2°tan3°...+tan88°tan89°
问题描述:
tan1°tan2°+tan2°tan3°...+tan88°tan89°
答
∵tan(α-β)=(tanα-tanβ)/(1+tanαtanβ)∴tanαtanβ=(tanα-tanβ)/tan(α-β) - 1∴tan1°tan2°+tan2°tan3°...+tan88°tan89°=(tan2°-tan1°)/tan1°+(tan3°-tan2°)/tan1°+...+(tan89°-tan88°)/tan1...