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=?
sina/cosa+cosa/sina=9/4=1/sinacosa ,(sinacosa)^2=16/81
tan^2+1/sinacosa+1/tan^2a=sin^2a/cos^2a+9/4+cos^2a/sin^2a
=9/4+(sin^4a+cos^4a)/sin^2acos^2a
=9/4+[(sin^2a+cos^2a)^2-2(sinacosa)^2]/(sinacosa)^2
=9/4+(1-32/81)81/16
=9/4+49/16=85/16
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)
已知1-tana/1+tana=根号5,则tan(4/派-a)=
(tana-1)/(tana+1)=3,则tan(a-派/4)=
已知1-tanA/1+tanA=2+根号3,则tan(π/4+A)是
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
怎么证tan(45+a)=(1+tana)/(1-tana).
化简tana+(1+tana)tan(π/4-a)
化简:tana+tan(派/2+a)+4tana/1-tanatana
tan(a+b)-1/tana=2*tana 请化简!
求证tan(A/2)-{1/(tanA/2)}=-2/tanA
求证:sin2A/2cosA(1+tan*tanA/2)=tanA
求证tan(a/2)-1/(tana/2)=-2/tana
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