Question

9、已知sinθ=3/5，θ∈(π/2，π)，tanφ=1/2，求tan(θ+φ)，tan(θ-φ)的值

Answer 1

解：∵sinθ=3/5，θ∈(π/2，π)
∴cosθ=-√(1-sin²θ)=-4/5
==>tanθ=sinθ/cosθ=-3/4
∵tanφ=1/2
∴tan(θ+φ)=(tanθ+tanφ)/(1-tanθtanφ)
=((-3/4)+(1/2))/(1-(-3/4)(1/2))
=-2/11；
tan(θ-φ)=(tanθ-tanφ)/(1+tanθtanφ)
=((-3/4)-(1/2))/(1+(-3/4)(1/2))
=-2。.