Question

已知tanA=112,求下列2式的值.（1）sina-3cosa/sina+cosa(2)sin的平方+sinacosa

Answer 1

tanA=112,则 (cosA)^2 = 1/[1+(tanA)^2] = 1/12545

(1)（sina-3cosa）/（sina+cosa） 分子分母同除以cosx

=(tanx - 3)/(tanx + 1) = 109/113

(2) (sinA)^2 +sinAcosA+2 = (cosA)^2 [(tanA)^2 + tanA] + 2 = (1/12545)*[112^2 + 112]

= 12656/12545