若0<α<π/2,π<β<3π/2,且tanα=1/7,tanβ=3/4,则α+β=?
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若0<α<π/2,π<β<3π/2,且tanα=1/7,tanβ=3/4,则α+β=?
若0<α<π/2,π<β<3π/2,且tanα=1/7,tanβ=3/4,则α+β=?
若0<α<π/2,π<β<3π/2,且tanα=1/7,tanβ=3/4,则α+β=?
tanα=1/7,tanβ=3/4
tan(α+β)=(tanα+tanβ)/(1-tanαtanβ)
=(1/7+3/4)/(1-3/4*1/7)
=1
又0<α<π/2,π<β<3π/2
所以π<α+β<2π
所以α+β=5π/4
0<α<π/2,π<β<3π/2
π<α+β<2π
tan(α+β)=(tanα+tanβ)/(1-tanαtanβ)=1,
∴α+β=5π/4