`sin(pi/2+theta)*cos((3pi)/2+theta)*tan((5pi)/2+theta)*cot((7pi)/2+theta)`

If tan theta = 3/4 and theta is not in the first quadrant then (sin (pi//2 + theta)-cot(pi-theta))/(tan(3 pi//2 - theta)-cos (3pi//2+theta))=

If tan theta=(3)/(4) and theta is not in first quadrant sin(sin((pi)/(2)+theta)-cot(pi-theta))/(tan((3 pi)/(2)-theta)-cos((3 pi)/(2)+theta))=

sin""((pi)/2+ theta ) * cos (pi- theta ) * cot ""((3pi)/(2)+ theta ) - sin""((pi)/2- theta ) * sin""((3pi)/2 - theta ) * cot ""((pi)/2 + theta ) =

Prove that (sin(3pi-theta)cos(theta- pi/2)tan((3pi)/(2)-theta))/(cosec((13pi)/(2)+theta)sec(3pi+theta)cot(theta- pi/2))=cos^(4)theta

sin((pi)/(2)+theta)+sin((3 pi)/(2)+theta)+sin((5 pi)/(2)+theta)+sin((7 pi)/(2)+theta) is equal to

Prove that (cos (pi-theta).cot(pi/2 +theta).cos(-theta))/(tan(pi+theta).tan((3pi)/(2)+theta).sin(2pi-theta))=cos theta

cos ((3pi)/(2)+theta) cos (2pi+theta) xx [ cot ((3pi)/(2) -theta) + cot (2pi + theta) ] = ?

x cot((pi)/(2)+theta)+tan((pi)/(2)+theta)sin theta+csc((pi)/(2)+theta)=0

(cosec(2pi+theta)cos(2pi+theta)tan(pi/2+theta))/(sec(pi/2+theta)costhetacot(pi+theta))