`[sin(pi/2-x) + sin(pi-x)]^2+[(cos((3pi)/2-x)+cos(2pi-x)]^2=`

Prove that: (a) (cos(pi+x) cos(-x))/(sin(pi-x)cos(pi/2+x))=cot^(2)x (b) cos((3pi)/2 + x)cos(2pi+x){cot ((3pi)/2-x)+cot(2pi+x)}=1

If the expression cos(x - 3pi/2) + sin(3pi/2+x) + sin(32pi+x) - 18 cos(19pi-x)+cos(56pi+x) - 9sin(x+17pi) is expressed in the form of a sinx + bcos x , then a+b is equal to

(cos(pi-x)cos(-x))/(sin(pi-x)cos(pi/2+x))=cot^(2)x

(cos (pi + x) cos (-x)) / (sin (pi-x) cos ((pi) / (2) -x)) = cot ^ (2) x

(cos(pi-x)cos(-x))/(sin(pi-x)cos(pi/2+x))=cot^(2)x

sin m((pi)/(2)+x)+cos n((pi)/(2)-x)

(cos (pi+x).cos(-x))/(sin(pi-x).cos(pi/2+x)) = cot^2x

Prove that (cos(pi+x)cos(-x))/(sin(pi-x)cos((pi)/(2)+x))=cot^(2)x