Question

知：tan（π+α）=-1/3,tan（α+β）=[sin2（π/2-α）+4（cosα）^2]/[10(cosα)^2

Answer 1

tan（π+α）=-1/3

tanα=-1/3

tan（α+β）=[sin2（π/2-α）+4cos²α]/[10cos²α-sin2α]

=(sin2α+4cos²α)/(10cos²α-sin2α)

=(2sinαcosα+4cos²α)/(10cos²α-2sinαcosα)

=(sinα+2cosα)/(5cosα-sinα)

=(tanα+2)/(5-tanα)

=5/16

tan（α+β）=(tanα+tanβ)/(1-tanαtanβ)

5/16=(-1/3+tanβ)/[1-(-1/3)tanβ]

tanβ=31/43