已知sinx=2√5/5,求tan(π+x)+sin(5π/2+x)/cos(5π/2-x)

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已知sinx=2√5/5,求tan(π+x)+sin(5π/2+x)/cos(5π/2-x)
已知sinx=2√5/5,求tan(π+x)+sin(5π/2+x)/cos(5π/2-x)

已知sinx=2√5/5,求tan(π+x)+sin(5π/2+x)/cos(5π/2-x)
解答:
∵ sinx=2√5/5,
∴ cos²x=1-sin²x=1-4/5=1/5
∴ cosx=±√5/5,
tan(π+x)+sin(5π/2+x)/cos(5π/2-x)
=tanx+ (cosx)/sinx
=sinx/cosx+cosx/sinx
=(sin²x+cos²x)/(sinxcosx)
=1/(sinxcosx)
=1/[(2√5/5)*(±√5/5)]
=±1/(2/5)
=±5/2