Question

已知sinθ=3/5,θ是第二象限角.tanφ=1/2,求tan(θ+φ)和tan(θ-φ)

Answer 1

tanφ=1/2 ,

sinθ=3/5 ,1-cos^2θ=9/25,cosθ=-4/5

tanθ=-3/4

tan(θ+φ)=(tanθ+tanφ)/(1-tanθ*tanφ)

=(-3/4+1/2)/(1+3/4*1/2)

=(-1/4)/(11/8)=-8/44

tan(θ-φ))=(tanθ-tanφ)/(1+tanθ*tanφ)

=(-3/4-1/2)/(1-3/4*1/2)

=(-5/4)/(5/8)=-2