化简:[tan(45°+A)/(1-tan²(45°+A)]×[sinAcosA÷(1-2sin²A]
问题描述：
化简:[tan(45°+A)/(1-tan²(45°+A)]×[sinAcosA÷(1-2sin²A]
答
:[tan(45°+A)/(1-tan²(45°+A)]×[sinAcosA÷(1-2sin²A]=1/2*[2tan(45°+A)/(1-tan²(45°+A)]×1/2[2sinAcosA÷(1-2sin²A]=1/2tan(90°+2A)*1/2tan2A=1/4cot2A*tan2A=1/4