cos(π/2-x)cos(π/2+x)cot(π-x)/sin(3π/2+x)cos(3π+x)tan(π+x)
问题描述:
cos(π/2-x)cos(π/2+x)cot(π-x)/sin(3π/2+x)cos(3π+x)tan(π+x)
答
cos(π/2-x)cos(π/2+x)cot(π-x)/[sin(3π/2+x)cos(3π+x)tan(π+x)]=sinx*sin(-x)*(-cotx)/[sin(-π/2+x)*cos(π+x)*tanx]=sinx*sinx*cotx/[-cosx*(-cosx)*tanx]=sinx*cosx/(cosx*sinx)=1