tana+1/tana=9/4,则tan^2+1/sinacosa+1/tan^2a=?
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tana+1/tana=9/4,则tan^2+1/sinacosa+1/tan^2a=?
tana+1/tana=9/4,则tan^2+1/sinacosa+1/tan^2a=?
tana+1/tana=9/4,则tan^2+1/sinacosa+1/tan^2a=?
∵tana+1/tana=9/4
∴tana=1/2.或tana=4
∴tan^2a+1/sinacosa+1/tan^2a
=tan^2a+(sin^2a+cos^2a)/sinacosa+1/tan^2a
=tan^2a+tana+1/tana+1/tan^2a
=1/4+1/2+2+4
=6又3/4
或=16+4+1/4+1/16
=20又5/16
tana+1/tana=9/4,则tan^2+1/sinacosa+1/tan^2a=?
tana+1/tana=9/4,则tan^2+1/sinacosa+1/tan^2a=?
求证tan(a+π/4)=(1+tana)/(1-tana)
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若1-tanA/1+tanA=根号5,则tan(π/4+a)
1-tanA/1+tanA=√5,则tan(π/4+A)的值为
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tan(a+b)-1/tana=2*tana 请化简!
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若1-tanA / 1+tanA=4+√5,则tan(A-5/4 π)的值=?
(1-tanA)/(1+tanA)=4+根号5,则tan(A-5π/4)=?(解答一下)
已知(1-tanA)/(1+tanA)=4+根号5,则tan(π/4+a)的值等于