cos((3π/2)+x)cos(2π+x)[cot((3π/2)-x)+cot(2π+x)]

Question

Solve the problem

Answer 1

cos(3pi/2)=0=sin(2pi); sin(3pi/2)=-1; cos(2pi)=1.
Identities: sin(A+B)=sinAcosB+cosAsinB; cos(A+B)=cosAcosB-sinAsinB; cotX=cosX/sinX.
cos(3pi/2 + x)=sin(x); cos(2pi + x)=cos(x);
cot(3pi/2-x)=cos(3pi/2-x)/sin(3pi/2-x)=-sin(x)/(-cos(x))=tan(x);
cot(2pi+x)=cot(x).
The whole expression becomes:
sin(x)cos(x)(tan(x)+cot(x))=sin^2(x)+cos^2(x)=1.