Question

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

Answer 1

$displaystyle frac{sin( frac{pi}{2}+2 alpha )-cos(2 pi -2 alpha )+tg(2alpha- frac{3pi}{2}) }{ctg( pi +2alpha)} =\ \ \ = dfrac{cos2alpha-cos2alpha-ctg2alpha}{ctg2alpha} =-1.$