若cot(∏/2+α)=-3,则4sinα-cosα/(5cosα+3sinα)=?

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若cot(∏/2+α)=-3,则4sinα-cosα/(5cosα+3sinα)=?若cot(∏/2+α)=-3,则4sinα-cosα/(5cosα+3sinα)=?若cot(∏/2+α)=-3,则

若cot(∏/2+α)=-3,则4sinα-cosα/(5cosα+3sinα)=?
若cot(∏/2+α)=-3,则4sinα-cosα/(5cosα+3sinα)=?

若cot(∏/2+α)=-3,则4sinα-cosα/(5cosα+3sinα)=?
∵cot(π/2+α)=-tanα=-3
∴tanα=3
∴(4sinα-cosα)/(5cosα+3sinα)
=(4tanα-1)/(5+3tanα)
=(4*3-1)/(5+3*3)
=11/14