Question

( sin(П/2+2a)-cos(2П-2a)+tg(2a-3П/2) ) : ctg(П+2d ) = ???

Answer 1

$\displaystyle \frac{\sin( \frac{\pi}{2}+2 \alpha )-\cos(2 \pi -2 \alpha )+tg(2\alpha- \frac{3\pi}{2}) }{ctg( \pi +2\alpha)} =\\ \\ \\ = \dfrac{\cos2\alpha-\cos2\alpha-ctg2\alpha}{ctg2\alpha} =-1.$