tana-2分之1,tan(a-b)

(1+tana+cota)/(1+tan^2a+tana)-cota/(1+tan^2a)=(1+sina/cosa+cosa/sina)/(1+sina^2/cosa^2a+sina/cosa)-c

由tan(2A)=2tanA/(1-(tanA)^2)得1-(tanA)^2=2tanA/tan(2A)把A变成A/2即1-(tan(A/2))^2=2tan(A/2)/tanA①原式=[(tan(A

再答：

当1-tana分之2无意义时1-tana=0tana=45度tan(a+15°)-tan(a-15°)=tan(45°+15°)-tan30°=√3-√3/3=2√3/3

tanA=sinA/cosA=2sin(A/2)cos(A/2)/[cos²(A/2)-sin²(A/2)]【同除以cos²(A/2)】=2tan(A/2)/[1-tan

tan(π/4+A)=sin(π/4+A)/cos(π/4+A)=(sinπ/4*cosA+cosπ/4*sinA)/(cosπ/4*cosA-sinπ/4*sinA)=(tanπ/4*cosA+si

设a=A则(1/tanA/2-tanA/2)(1+tanAtanA/2)=[(1-tan²A/2)/tanA/2][1+2tan²A/2/(1-tan²A/2)]=[(1

tan(A-B)=(tanA-tanB)/(1+tanA*tanB)tan(A-B)/tanA+sin²C/sin²A=1左右移项得1-[(tanA-tanB)/(1+tanA*t

tan(π/4＋a)=[tan(π/4)＋tana]/[1－tan(π/4)tana]=(1＋tana)/(1－tana)=1/2解得：tana=－1/3

tana+(1+tana)tan(π/4-a)=tana+(1+tana)(1-tana)/(1+tana)=tana+1-tana=1

tan(a/2)-1/tan(a/2)=[tan²(a/2)-1]/tan(a/2)=-2[1-tan²(a/2)]/2tan(a/2)=-2/{2tan(a/2)/[1-tan&

结果:tana*tanb=1/2.过程也不复杂,把tana移项,然后展开tan(a+b),再全部通分,两边合并同类项.

tana+1/tana=3可化成sina/cosa+cosa/sina=3化简得sin^2a+cos^2a/sinacosa=3可得出sinacosa=1/3由此可得出sina+cosa=根号15/3

tanA=2tan(A/2)/[1-(tan(A/2))^2]-2/tanA=-2*[1-(tan(A/2))^2]/[2tan(A/2)]=[(tan(A/2))^2-1]/(tanA/2)tan(

证明：tan(a+b)=(tana+tanb)/(1-tana·tanb)∴tan(a+π/4)=[tana+tan(π/4)]/[1-tana·tan(π/4)]=(1+tana)/(1-tana)

tan(a/2)-1/(tana/2)=sin(a/2)/cos(a/2)-cos(a/2)/sin(a/2)通分=[sin²(a)-cos²(a/2)]/[sin(a/2)cos

分析法倒推tanr=-tan(a-r-a)=[tana-tan(a-r)]/[1+tana*tan(a-r)]tana*tanr=[tan^2a-tana*tan(a-r)]/[1+tana*tan(

∵tanA-1/tanA=2∴平方,tan²A-2+1/tan²A=4∴tan²A+1/tan²A=6

左边=1+sin^2a/cos^2a=(cos^2a+sin^2a)/cos^2a=1/cos^2a=sina/(sinacos^2a)=sina/(cosa×sinacosa)=tana/(sina

tan(45+A)=(tan45+tanA)/(1-tan45tanA)tan45=1tan(45+A)=(1+tanA)/(1-tanA)=1/(2+√3)=2-√3