中4數學.請列詳細既步驟,感激,thx

1. cos ( 90 - A ) * sin ( 360 - A ) + cos ( 180 - A ) * sin ( 90 - A )
= sin A * ( -sin A ) + ( -cos A ) * cos A
= -sin^2 A + ( -cos^2 A )
= -1

2. [ tan ( 270 + A ) * cos^2 ( 90 - A ) ] / [ cos^2 ( 180 + A ) * tan ( 90 - A ) ]
= { tan [ 360 - ( 90 - A ) ] * sin^2 A } / [ -cos^2 A * ( 1 / tan A ) ]
= ( - 1 / tan A ) * tan^2 A * tan A
= -tan^2 A

First, yo should know c
os(90-A)=sinA
sin(360-A)=-sinA
cos(180-A)=-cosA
sin(90-A)=cosA
cos(180+A)=-cosA
tan(270+A)=-tanA
tan(90-A)=1/tanA
tanA=sinA/cosA
then 1. cos(90-A)*sin(360-A)+cos(180-A)*sin(90-A)
=(sinA)(-sinA)+(-cosA)(cosA)
=-sin^2A-cos^2A
=-(sin^2A+cos^2A)
=-1

2. tan(270+A)*cos^2(90-A)/cos^2(180+A)*tan(90-A)
=-(1/tanA)*sin^2A/(-cos^2A)*(1/tanA)
=-tan^2A*(-1/tan^2A)
=1

cos(90-A)*sin(360-A)+cos(180-A)*sin(90-A)
=cosA*(-sinA)+(-cosA)*sinA
=-2cosAsinA

tan(270+A)*cos^2(90-A)/cos^2(180+A)*tan(90-A)
=(tanAcos^2A)/(cos^2AtanA)
=1

2008-02-24 16:06:47 補充：
SOR AR原來我計錯左=3=cos(90-A)*sin(360-A) cos(180-A)*sin(90-A)=sinA*(-sinA) (-cosA)*cosA=-sin^2A-cos^2A=-(sin^2A cos^2A)=-1tan(270 A)*cos^2(90-A)/cos^2(180 A)*tan(90-A) =[(1/tanA)(cos^2A)]/[(cos^2A)(1/tanA)=1