Question

cos(П/2 + х) * ctg П/4 * ctg(П/2 - L) / tg(П+L) * sin(2П+х)​

Answer 1

$\dfrac{cos\Big(\dfrac{\pi}{2}+x\Big)\cdot ctg\dfrac{\pi}{4}\cdot ctg\Big(\dfrac{\pi}{2}-L\Big)}{tg(\pi +L)\cdot sin(2\pi+x)}=\dfrac{-sinx\cdot 1\cdot tgL}{tgL\cdot sinx}=-1$