Помогите решить дз!!

$\sqrt{[ctg^2(x)-tg^2(x)]*cos(2x)}*tg(2x)=\\\\= \sqrt{[\frac{cos^2(x)}{sin^2(x)}-\frac{sin^2(x)}{cos^2(x)}]*cos(2x)}*tg(2x)=\\\\= \sqrt{ \frac{cos^4(x)-sin^4(x)}{sin^2(x)cos^2(x)}*cos(2x) }*tg(2x) =\\\\= \sqrt{4* \frac{[cos^2(x)-sin^2(x)]*[cos^2(x)+sin^2(x)]}{[2sin(x)cos(x)]^2}*cos(2x) }*tg(2x) =\\\\= \sqrt{4* \frac{cos^2(x)-sin^2(x)}{[sin(2x)]^2}*cos(2x) }*tg(2x) =\\\\= \sqrt{2^2* \frac{cos(2x)}{[sin(2x)]^2}*cos(2x) }*tg(2x) =\\\\= 2 \sqrt{ [\frac{cos(2x)}{sin(2x)}]^2}*tg(2x) =$

$=2 \sqrt{ctg^2(2x)}*tg(2x)=2*|ctg(x)|*tg(x)=2*(-ctg(2x))*tg(2x)=\\\\= 2*(-1)=-2,\ if\ \frac{\pi}{4} \ \textless \ x \ \textless \ \frac{\pi}{2}$

P.S. $if\ \ \ \frac{\pi}{4} \ \textless \ x \ \textless \ \frac{\pi}{2},\ then\ \ \ \frac{\pi}{2} \ \textless \ 2x \ \textless \pi \\\\ and\ \ \ ctg(2x)\ \textless \ 0$