sin(30°+α)×tan(135°+α)×tan(135°-α)×1/cos(60°-α)
问题描述：
sin(30°+α)×tan(135°+α)×tan(135°-α)×1/cos(60°-α)
答
sin(30°+α)×tan(135°+α)×tan(135°-α)×1/cos(60°-α)
= (sin 30°cos α +cos 30°sin α) ×{(tan135° +tanα) /(1-tan135°tanα)}×{(tan135° -tanα) /(1+tan135°tanα)}×1/(cos60°cos α+sin60°sinα)
= {1/2cos α+（根号3）/2sinα}×{(-1+tanα)/(1+tanα)}×{ (-1-tanα)/ (1-tanα) }× 1/{ (1/2cos α+(根号3）/2sinα
= {(-1+tanα)/(1+tanα)}×{ (-1-tanα)/ (1-tanα) }
=1
tan135°=tan(180°-45°)= -tan45°=-1