Question

1) sin2x\2cosx 2) tg^2x(1+cos2x) 3)cos^2-cos2x 4) cos2x-cos^2x\1-cos^2x

Answer 1

$\frac{Sin2x}{2Cosx} = \frac{2SinxCosx}{2Cosx}=Sinx\\\\tg ^{2}x(1+Cos2x) = \frac{Sin ^{2}x }{Cos ^{2}x }*2Cos ^{2} x=2Sin ^{2}x\\\\Cos ^{2} x-Cos2x=Cos ^{2}x-(Cos ^{2} x-Sin ^{2}x)=Cos ^{2} -Cos ^{2}x+Sin ^{2}x$$=Sin ^{2}x\\\\ \frac{Cos2x-Cos ^{2}x }{1-Cos ^{2}x }= \frac{Cos ^{2}x-Sin ^{2}x-Cos ^{2}x }{Sin ^{2} x} = \frac{-Sin ^{2}x }{Sin ^{2} x}=-1$

Answer 2

$1) \frac{sin2x}{2cosx} = \frac{2cosxsinx}{2cosx} = sinx$
$2) tg^2x(1+cos2x)= \frac{sin^2x}{cos^2x} * 2cos^2x = 2sin^2x$
$3)cos^2x-cos2x = cos^2x-cos^2x + sin^2x=sin^2x$
$4) \frac{cos2x-cos^2x}{1-cos^2x} = \frac{cos^2x - sin^2x-cos^2x}{sin^2x} = -1$