2求证:tanA +tanB + tanC = tanA.tanB.tanC 3在非直角三角形中,求证:tan A+tan B+tan C=tan Atan Btan C. 4在非直角三角形中 求证 tanA+tanB+tanc=tanAtanBtanC 5在斜三角形ABC中,求证:tanA+tanB+tanC=tanA*tanB*tanC.反馈 收藏
即 (tanA+tanB)/(1-tanAtanB)=-tanC,去分母得 tanA+tanB=-tanC+tanAtanBtanC,整理得 tanAtanBtanC=tanA+tanB+tanC。在锐角三ΔABC中sinA=sin(B+C)=sinBcosC+cosBsinC由已知sinBcosC+cosBsinC=2 sinBsinCtanB+tanC=2tanBtanC (1)tanA=-tan(B+C)=(tanB+tanC)/(tanBtanC-1)tanA=...
∴9tanAtanB+tanBtanC+tanCtanA= +( + ) , = + , ≥ + , =( + )[mn+(1﹣mn)], =9+4+ + , ≥13+2 =13+12=25,当且仅当 = ,即m=n= 时取等号, 故最小值为25, 所以答案是:25 .【考点精析】本题主要考查了基本不等式在最值问题中的应用的相关知识点,需要掌握用基本不等式求最值时...
在斜△ABC中,tanAtanB+tanBtanCtanCtanA=tanB(tanA+tanC)tanAtanC=sinBcosB(sinAcosA+sinCcosC)sinAsinCcosAcosC=sinB(sinAcosC+cosAsinC)sinAsinCcosB=sinBsin(A+C)sinAsinCcosB=sin2BsinAsinCcosB=b2accosB=2b22a... 运用切化弦和两角和的正弦公式及诱导公式,再由正弦定理、余弦定理,即可得到. 本题考点:...
在斜△ABC中，tanAtanB+tanBtanCtanCtanA=tanB(tanA+tanC)tanAtanC=sinBcosB(sinAcosA+sinCcosC)sinAsinCcosAcosC=sinB(sinAcosC+cosAsinC)sinAsinCcosB=sinBsin(A+C)sinAsinCcosB=sin2BsinAsinCcosB=b2accosB=2b22a...
锐角三角形ABC中,求证:tanA+tanB+tanC=tanAtanBtanC证:A+B+C+180°,C=180°-(A+B)tanC=tan[180°-(A+B)]=-tan(A+B)∴tanC=-[(tanA+tanB)/(1-tanAtanB)]∴tanC×(1-tanAtanB)=-(tanA+tanB)∴tanC-tanAtanBtanC=-(tanA+tanB)∴tanA+tanB+tanC=tanAtanBtanCtanC=tan(π-A-B)=-tan(A+B)...
tanAtanBtanC=tanA+tanB+tanC 证明：A+B+C=π ∵A+B＝π-C，∴tan(A+B)=tan(π-C)(tanA+tanB)/(1-tanAtanB)=-tanC,tanA+tanB=-tanC+tanAtanBtanC ∴tanA+tanB+tanC=tanAtanBtanC.因为
(B+C) =-(tanH+tanθ)/(1-tanBtanC) ②, 则, 由f(a)B=SωaC=2amABsanC可得 tanAtanBcosC=(2(tant)tan(1-t)^2)/(1-tan2tanα), 令,由A,B,C为锐角, 可得,,f(a)C0, 由式得1-2anBsanC0,解得R1, , 又1/(e^2)+1/R=(1/2-1/2)^2-1/4, 由R1得,, 因此tanAtanBtanC的最小...
tanAtanB+tanBtanC+tanCtanA=tanAtanB+(tanA+tanB)ctan(B+A) (by A+B+C=90°)=(sinAsinB)/(cosAcosB)+[(sinA/cosA)+(sinB/cosB)]cos(A+B)/sin(A+B)=(sinAsinB)/(cosAcosB)+[(sinAcosB+sinB+cosA)/(cosAcosB)]cos(A+B)/sin(A+B)=(sinAsinB)/(cosAcosB)+cos(A+B)/(cosAcosB)=1 解...
tanAtanB=tanAtanC+tanBtanC sinA/cosA * sinB/cosB =sinA/cosA * sinC/cosC + sinB/cosB * sinC/cosC sinAsinBcosC=sinAsinCcosB+sinBsinCcosA 根据正弦定理，a/sinA=b/sinB=c/sinC=2R 代入，得 sinA=a/(2R),sinB=b/(2R),sinC=c/(2R)abcosC/(4R^2)=accosB/(4R^2)+bccosA/(...