Question

详解'谢谢

Answer 1

解：
∵sinα=3/5 , π/2﹤α﹤π
∴cosα﹤0
∵sin²α+cos²α=1
∴cos²α=1-sin²α
cosα=-√(1-sin²α)=-√[1-(3/5)²]=-4/5
∴tanα=sinα/cosα=(3/5)/(-4/5)=-3/4
∵tan(π-β)=1/2
∴-tanβ=1/2
tanβ=-1/2
∴tan2β=2tanβ/(1-tan²β)
=2×(-1/2)/[1-(-1/2)²]
=-1/(3/4)
=-4/3

tan(α-2β)
=(tanα-tan2β)/(1+tanα·tan2β)
=[-3/4-(-4/3)]/[1+(-3/4)×(-4/3)]
=(-3/4+4/3)/(1+1)
=(7/12)/2
=7/24
正确答案应该选 D

Answer 2

sinα=3/5;
tanα = -3/4(п/2,п)
tan(п-β )=1/2
tan(2п-2β )=1/2*2/(1-1/2*1/2)=4/3
tan(-2β )=4/3
tan(α-2β )=[tanα+tan(-2β)] / [1 - tanαtan(-2β)]=(-3/4+4/3)/[1-(-3/4)×(4/3)] =7/24