(1-tan^4A)cos^2A tan^2A=1

sin(π/4+a)=cos[π/2-(π/4+a)]=cos(π/4-a)所以分母=2[sin(π/4-a)/cos(π/4-a)]*cos²(π/4-a)=2sin(π/4-a)cos(

因为cot(a/2)-tan(a/2)=cos(a/x2)/sin(a/2)-sin(a/2)/cos(a/2)(通分)=[(cos(a/2))^2-(sin(a/2))^2]/[cos(a/2)si

tan(π/4+a)=[tanπ/4+tana)/[1-tanπ/4*tana]=2(1+tana)/(1-tana)=2tana=1/31/(2sinacosa+cos²a)=(sin&#

tan[b－(π/4)]=[tanb－tan(π/4)]/[1＋tan(π/4)tanb]=[tanb－1]/[1＋tanb]=1/4则：tanb=5/3,又：tan(a＋b)=2/5则：tana=t

(2cos²a-1)/[2tan(π/4-a)*sin²(π/4+a)]=cos2a/[2tan(π/4-a)*cos²(π/4-a)]=cos2a/[2sin(π/4-

cos^4a-sin^4a=(cos^2a+sin^2a)(cos^2a-sin^2a)=(cosa+sina)(cosa-sina)=(cosa+sina)(cosa-sina)=cos^2a(1-

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

cos(a/2)不等于0.tan(a/2)=sin(a/2)/cos(a/2)=2sin(a/2)cos(a/2)/{2[cos(a/2)]^2}=sin(a)/[1+cos(a)]当sin(a/2)

右边=2sin(a/2)cos(a/2)/(1+2cos(a/2)平方-1)=sin(a/2)/cos(a/2)=左边

第一问的方法是将1拆成sin²a+cos²a,然后就能算了第二问用到常用的倍角公式：cos2θ=cos²a-sin²a=2cos²a-1=1-2sin

我觉得应该是tana+1/tana=(sin²a+cos²a)/(sinacosa)=1/(sinacosa)你那个,举个例子：a=30°左边=√3+√3/3右边=1/4+（3/4

tana/2=sin(a/2)/cos(a/2).分子分母同乘以cos(a/2),得=sin(a/2)*cos(a/2)/cos(a/2)*cos(a/2)=（1/2）sina/(1/2)(1+cos

tana(-cosa)sin²a/(tanasin²a)=0.5cosa=-0.5sin²+cos²a=1所以sina=±√3/2所以tana=sina/cos

(4sina-2cosa)/(5cosa+3sina)=(4tana-2)/(5+3tana)=(-4/3-2)/[5+3(-1/3)]=(-10/3)/4=-10/12=-5/6

cos4a/cos2a

=(2cos2a-1)2=[2tan(π/4-a)*cos2(π/4-a)]=cos22a/[2sin(π/4-a)cos(π/4-a)]=cos22a/sin(π/2-2a)=cos22a/cos2

=（1+sin^2a/cos^2a）cos^2a=sin^2a+cos^2a=1

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

(2cos^2A-1)/2tan(π/4-A)cos^2(π/4-A)=(cos2A)/[2sin(π/4-A)cos(π/4-A)]=(cos2A)/sin(π/2-2A)=(cos2A)/(cos