Question

1/tg^2x-2cos2x/1+sin(2x+1,5pi)​

Answer 1

$frac{1}{tg^{2} x}-frac{2Cos2x}{1+Sin(2x+1,5pi)}=frac{1}{tg^{2}x }-frac{2Cos2x}{1-Cos2x}=frac{Cos^{2}x }{Sin^{2}x }-frac{2Cos2x}{2Sin^{2}x } =frac{Cos^{2}x-Cos^{2}x+Sin^{2}x}{Sin^{2}x }=frac{Sin^{2}x }{Sin^{2}x }=1$