已知tan(π a)=-1 2

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(b-π/3)=1/3∴tan(π/3-b)=-1/3∵tan(a+b)=3/5∴tan(a+π/3)=tan[(a+b)+(π/3-b)]=[tan(a+b)+tan(π/3-b)]/[1-

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

若a∈(0,π),cosa=3/5sina=4/5f(a)=sin(π+a)tan(π+a)cos(-a)/cos(3π-a)tan(2π-a)=-sinatanacosa/(cosatana)=-s

(1)f(a)=-cosa(2)f(a)=1/5用口诀“奇变偶不变,符号看象限”!

我发现题目是存在问题的,a为第二象限角,所以π/2＜a＜π,所以-π＜a-3π/2＜-π/2因为tana的周期是π,由其函数图象可知,a-3π/2大于0不可能是负的.所以第二问是存在问题的.第一问慢慢

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

∵（sinα+cosα）2=sin2α+2sinαcosα+cos2α，cos2α=cos2α-sin2α∴(sinα+cosα)2cos2α=sin 2α+2sinαcosα+cos&nb

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]=

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(

f(a)=sinacosacotatana/(-sina)所以f(a)=-cosa

由Sin（a-（2n+1）π/2）=3/5化简得cosa=3/5或-3/5故易知tana=4/3或-4/3所以tana+1/tana=25/12或-25/12

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

f(a)=sin(π-a)cos(2π-a)tan(-a+3π/2)/tan(π/2+a)sin(-π-a)=sinacosatan(π/2-a)/[cot(-a)sina=cosacota/(-co