已知sin(π+α)=-3/5,α∈(0,π/2),求tan(π/2-α)的值.

问题描述:

已知sin(π+α)=-3/5,α∈(0,π/2),求tan(π/2-α)的值.

sin(π+a)=-3/5.又sin(π+a)=-sina.∴sina=3/5.又0<a<π/2.∴cosa=4/5.∴cota=(cosa)/(sina)=4/3.∴tan[(π/2)-a]=cota=4/3.

∵sin(π+α)=-3/5,
∴sinα=3/5,
又α∈(0,π/2),
∴cosα=4/5
tan(π/2-α)
=sin(π/2-α)/cos(π/2-α)
=cosα/sinα
=(4/5)/(3/5)
=4/3.