已知cos(π/2+α)=-2(√5)/5,且α∈(π/2,π),(1)求tanα的值;
(2)求[sin(π/2+α)-sin(π-α)]/[cos(π/2-α)-cos(π+α)]的值
(1).cos(π/2+α)=-sinα=-2(√5)/5,故sinα=2(√5)/5,α∈(π/2,π),
∴tanα=-(2√5)/√(25-20)=-2
(2)[sin(π/2+α)-sin(π-α)]/[cos(π/2-α)-cos(π+α)]=(cosα-sinα)/(sinα+cosα)
=(1-tanα)/(tanα+1)=(1+2)/(-2+1)=-3.