数学三角函数

(1)tanb=tana+seca => sinb/cosb=(sina+1)/cosa => sinb*cosa=sina*cosb+cosb => sin(b-a)=cosb => b-a+b=π/2+2kπ (k=0,-1,+1,-2,+2,-3,+3……) => 2b=a+π/2+2kπ => cos2b=-sina => cos2a-2cos2b=1-2(sina)~2+sina 所以当且仅当sina=1/2或0时，cos2a-2cos2b=1
而当sina≠1/2且sina≠0时，cos2a-2cos2b≠1
(2)2tana=3tanb => 2sina/cosa=3sinb/cosb => sina*cosb=3sin(a-b),cosa*sinb=2sin(a-b);
欲证tan(a-b)=(sin2b)/(5-cosb)
只需(sina*cosb-sinb*cosa)/(cosa*cosb+sina*sinb)=2sinb*cosb/[5-(cosb)~2+(sinb)~2]
只需sin(a-b)/(cosa*cosb+sina*sinb)=2sinb*cosb/[5-(cosb)~2+(sinb)~2]
只需2cosa*sinb*(cosb)~2+2sina*cosb*(sinb)~2=sin(a-b)*[5-(cosb)~2+(sinb)~2]
只需4sin(a-b)*(cosb)~2+6sin(a-b)*(sinb)~2=5sin(a-b)-sin(a-b)*(cosb)~2+sin(a-b)*(sinb)~2
只需5sin(a-b)*(cosb)~2+5sin(a-b)*(sinb)~2=5sin(a-b)
因为(cosb)~2+(sinb)~2=1
所以可证：若2tana=3tanb,则tan(a-b)=(sin2b)/(5-cos2b)

第一问没看懂啊……
第二问：
证明：
tan(a-b)=(tana-tanb)/(1+tana*tanb)
因为2tana=3tanb
所以tan(a-b)=(0.5*t