已知tan(a-π 4)=1 2,则sina cosa sina-cosa=

因为cotB=1/3所以tanB=3tan（A+B）=(tanA+tanB)/(1-tanAtanB)=（4+3）/（1-4*3）=-7/11

用和角来算因为tan（a+β）=2/5,tan（β-π/4）=1/4a+β-(β-π/4)=a+π/4所以tan（a+π/4）=[tan（a+β）-tan（β-π/4）]/[1+tan（a+β)tan

tan(9π/4+a)*tan(3π/4+a)=tan(2π+π/4+a)*tan(π-π/4+a)=tan(a+π/4)*tan(a-π/4)=(1+tana)/(1-tana)*(1-tana)/

tan(a-π/4)=(tana-tanπ/4)/（1+tana*tanπ/4）=（2-1）/（1+2*1）=2/3

tan(a-4/π)=(tana-tanπ/4)/(1+tanatanπ/4)=(tana-1)/(1+tana)=2,所以tana=-3而1/(2sinacosa+cos^2a)=(sin^2a+c

tan（a+π/4）=（tana+tanπ/4）/1-tana*tanπ/4=（2+1）/1-2*1=-3

tan(b+π/4)=1/4[tanb+tan(π/4)]/[1-tanbtan(π/4)]=1/4(tanb+1)/(1-tanb)=1/45tanb=-3tanb=-3/5tan(a+b)=1/5

1.∵tan(a/2)=2∴tana=[2tan(a/2)]/{1-[tan(a/2)]^2}=(2×2)/(1-2^2)=-4/3∴tan(a+π/4)=[tana+tan(π/4)]/[1-tan

tan（π/4+a）=[tan(π/4)+tana]/[1-tan(π/4)tana]=(1+tana)/(1-tana)=2tana=1/3sinacosa+cos²a=(sinacosa

1.tan(a+π/4)=tan[(a+p)-(p-π/4)]=[tan(a+p)-tan(p-π/4)]/[1+tan(a+p)*tan(p-π/4)]=[2/5-1/4]/[1-2/5*1/4]=

cosa=-4/5sina=3/5tana=-3/4tan(π/4+a)=(tanπ/4+tana)/(1-tanπ/4tana)=(1+tana)/(1-tana)=1/7

左边=(tana-tanb)/(-1/tana+cotb)=(tana-tanb)/(-1/tana+1/tanb)上下乘tanatanb=tanatanb(tana-tanb)/(tana-tanb

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

tana=[sin2a]/[cos2a＋1]=A/(B＋1)tan(a＋π/4)=[tana＋tan(π/4)]/[1－tanatan(π/4)]=[1＋tana]/[1－tanA]=(A＋B＋1)/

∵tan（a+π4）=tana+11−tana=13∴tana=-12因此，(sina−cosa)2cos2a=sin2a−2sinacosa+cos2acos2a−sin2a分子分母都除以cos2a

tan（a+pai/4)=x用未知数x表示,比较简便tan（a+b）=tan[(a+pi/4)+(b-pi/4)]=[tan(a+pi/4)+tan(b-pi/a)]/[1-tan(a+pi/4)*t

tan(π/4+a)=(tanπ/4+tana)/(1-tanπ/4*tana)=(tana+1)/(1-tana)=3tana+1=3-3tanatana=1/2sin2a-2cos^2a-1=si

tan(π/4+a)=（1+tana）/（1-tana）=-1/2tana=-3sina=3根号10/10[sin2a-2(cosa)^2]/1+tana=[sin2a-2+2(sina)^2]/1+

解tan(-a-4π/3)=tan[-(a+4π/3)]=-tan(a+4π/3)=-tan(a+π+π/3)=-tan(a+π/3)=-5∴tan(a+π/3)=5

因为cos2a=B,即cos^2a-sin^2a=B(^2表示平方）所以1-2sin^2a=B化简得：sina=(|2-2B)/2(|表示根号）(1)由sin2a=A得sina=A/2cosa(2)(