方程x^2+3ax+3a+1=0 (a>2)两根tanα,tanβ,且α,β属于（-π/2,π/2）,则α+β=

x^2+3ax+3a+1=0 (a>2)两根tanα,tanβ
tanα+tanβ=-3a
tanαtanβ=3a+1
上式+下式得
tanα+tanβ+tanαtanβ=1
tan(α+β)
=(tanα+tanβ)/(1-tanαtanβ)
=(1-tanαtanβ)/(1-tanαtanβ)
=1
∵α,β属于（-π/2,π/2）
∴α+β=π/4
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