题目内容
解析：原式=tan［(2α+β)+(α-β)］=tan3α.
答案：tan3α
A. B. C. D.
A.p+q+1=0 B.p-q+1=0
C.p+q-1=0 D.p-q-1=0
(2).
(2)cos(60°-α)-sin(60°-α).
tanα-tanβ=tan(α-β)·(_______________).
A.cos80°cos20°+sin80°sin20°=
B.sin13°cos17°-cos13°sin17°=
C.sin70°cos25°-cos65°sin20°=
D.sin140°cos20°+sin50°sin20°=
=______________.
=______________.=______________.
=__________________.
=
,则
等于( )
B.
C.
D.
-θ)是方程x2+px+q=0的两个根，则p、q间关系是( )
tan20°tan40°;
.
cos(60°-α)-sin(60°-α).
