化简：cos(π-α)·sin(-π-α) / cos(11/2π-β)·tan(3π+β)

题目

化简：cos(π-α)·sin(-π-α) / cos(11/2π-β)·tan(3π+β)
RT、
quickly~

答案

cos(π-α)sin(-π-α) / cos(11/2π-β)tan(3π+β)
=cos(π-α)sin[-(π+α)] / cos[4π+(3/2π-β)]tan[2π+(π+β)]
=-cosαsinα / (-sinβ)tanβ
=cosαsinαcosβ / (sinβ)^2