Question

sin 2(t-3П/2)cos 2(2П-t) : tg 2(t-п/2)cos 2(t-3п/2)

Answer 1

$\frac{sin^2(t-\frac{3\pi}{2})\, cos^2(2\pi -t)}{tg^2(t-\frac{\pi }{2})\, cos^2(t-\frac{3\pi }{2})}=\frac{cos^2t\cdot cos^2t}{ctg^2t\cdot sin^2t}=\frac{cos^4t}{\frac{cos^2t}{sin^2t}\cdot sin^2t}=cos^2t$