1、3.1.3两角和与差的正切课时过关能力提升1.已知tan(+)=25,tan+4=322,则tan-4等于()A.15B.1318C.14D.1322解析:tan-4=tan(+)-+4=tan(+)-tan+41+tan(+)tan+4=25-3221+25322=14.答案:C2.已知0,2,满足tan(+)=324,sin =13,则tan 等于()A.23B.4211C.3211D.324解析:由已知可得cos =1-sin2=223,从而tan =24,于是tan =tan(+)-=342-241+34224=4211.答案:B3.在ABC中,已知tan A,tan B是方程3x2+
2、8x-1=0的两根,则tan C等于()A.2B.-2C.4D.-4答案:A4.在ABC中,C=23,3tan A+3tan B=23,则tan Atan B的值为()A.14B.13C.12D.53解析:由C=23得A+B=3,于是tan(A+B)=tanA+tanB1-tanAtanB=3.即2331-tanAtanB=3,因此tan Atan B=13.答案:B5.在ABC中,tan A=12,cos B=31010,则tan C等于()A.-1B.1C.3D.-2解析:cos B=31010,且0B,sin B=1-cos2B=1010.tan B=13,tan C=-tanA+B=-
3、tanA+tanB1-tanAtanB=-12+131-1213=-1.故选A.答案:A6.设tan 和tan 是关于x的方程mx2+(2m-3)x+(m-2)=0的两根,则tan(+)的最小值是()A.154B.34C.-34D.不确定解析:依题意tan +tan =-2m-3m,tan tan =m-2m,于是tan(+)=-2m-3m1-m-2m=32-m.又方程有两根,所以=(2m-3)2-4m(m-2)0,即m94,因此32-m-34,即tan(+)的最小值为-34.答案:C7.已知sin 2=3542,tan(-)=12,则tan(+)=.解析:sin 2=35,cos 2=45.
4、又42,22,cos 2=-45,tan 2=-34.又tan(-)=12,tan(+)=tan2-(-)=tan2-tan(-)1+tan2tan(-)=-34-121-3412=-2.答案:-28.已知tan4+=2,则12sincos+cos2的值为.答案:239.在ABC中,若(1+cot A)(1+cot C)=2,则log2sin B=.解析:由(1+cot A)(1+cot C)=2,得tanA+1tanAtanC+1tanC=2,(tan A+1)(tan C+1)=2tan Atan C.1+tan A+tan C=tan Atan C.tan(A+C)=-1.又A,B,C是
5、ABC的内角,A+C=34.B=4.sin B=22.log2sin B=log222=-12.答案:-1210.已知为第二象限的角,sin =35,为第一象限的角,cos =513,求tan(2-)的值.解:为第二象限的角,且sin =35,cos =-45,tan =-34.为第一象限的角,且cos =513,sin =1213,tan =125.tan(-)=tan-tan1+tantan=-34-1251+-34125=6316.tan(2-)=tan+(-)=tan+tan(-)1-tantan(-)=-34+63161-346316=204253.11.如图,在矩形ABCD中,AB=a,BC=2a,在BC上取一点P,使AB+BP=PD,求tan APD的值.解:由AB+BP=PD,得a+BP=a2+(2a-BP)2,解得BP=2a3.设APB=,DPC=,则tan =ABBP=32,tan =CDPC=34.从而tan(+)=tan+tan1-tantan=-18.APD+(+)=,tanAPD=tan-(+)=-tan(+)=18.
