Question

$\sqrt{(ctg^2a-tg^2a)*cos2a} *tg2a$

Answer 1

Ответ:

2

Объяснение:

$\sqrt{(ctg^2a-tg^2a)*cos2a} *tg2a=\sqrt{(cos^2a/sin^2a-sin^2a/cos^2a)*cos2a}*tg2a=\sqrt{((cos^4a-sin^4a)/sin^2a*cos^2a)*cos2a}*tg2a=$

$\sqrt{\frac{(cos^2a-sin^2a)*(cos^2a+sin^2a)}{sin^2a*cos^2a}*cos2a}*tg2a=\\\sqrt{\frac{4*(cos^2a-sin^2a)}{4*sin^2a*cos^2a}*cos2a}*tg2a=\\\sqrt{4*\frac{cos2a}{sin^22a}*cos2a}*tg2a=2\sqrt{\frac{cos^22a}{sin^22a}}*tg2a=2*ctg2a*tg2a=2$