Question

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Answer 1

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

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

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