已知sinθ=3/5,θ∈(π/2,π),tanμ

1个回答

sinθ=3/5,θ∈(π/2,π),cosθ=-4/5 tanθ=-3/4 tan(θ+μ)=(tanθ+tanμ)/(1-tanθtanμ)=(-3/4+1/2)/(1+3/4X1/2)=-2/11 tan(θ—μ)=(tanθ-tanμ)/(1+tanθtanμ)=(-3/4-1/2)/(1-3/4X1/2)=-2

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