已知tanα=3分之2,求sinαcosα分之1=?
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tanα=2/3
1/(sinαcosα)
=2/(2sinαcosα)
=2/sin(2α)
=2/(2tanα/(1+(tanα)^2))
=(1+(tanα)^2)/tanα
=(1+(2/3)^2)/(2/3)
=(13/9)/(2/3)
=13/6
1/(sinαcosα)
=2/(2sinαcosα)
=2/sin(2α)
=2/(2tanα/(1+(tanα)^2))
=(1+(tanα)^2)/tanα
=(1+(2/3)^2)/(2/3)
=(13/9)/(2/3)
=13/6
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