Question

тригонометрия. 63-66.

Answer 1

$63)fracpi4 < a < fracpi2\ sqrt{(ctg^2a-tg^2a)cos2a}*tg2a = sqrt{frac{cos^22a}{sin^2acos^2a}}*tg2a = frac{cos2a}{sinacos}*frac{2sinacosa}{cos2a} = 2\64) fracpi2 < a < frac34pi = > sina>0, cosa, tga, ctg < 0\ sqrt{(ctga-tga)*2ctg2a} * tg2a +2 = sqrt{frac{cos2a}{sinacosa}*2frac{cos2a}{sin2a}}}*tg2a + 2 = sqrt{frac{cos^22a}{sin^2acos^2a}}*tg2a + 2 = -frac{cos2a}{sinacosa}*frac{2sinacosa}{cos2a} + 2 = -2+2 = 0$

$65)frac32pi<a<2pi => cosa > 0, sina, tga, ctga < 0\sqrt{sin^2a(1-ctga)+cos^2a(1-tga)} = sqrt{sina(sina-cosa) - cosa(sina-cosa)} = \ </p><p>=sqrt{(sina-cosa)(sina-cosa)} = sina-cosa$

$66)pi < b < frac32pi, sinb, cosb < 0, tgb, ctgb > 0\sqrt{cos^2b(1+tgb) + sin^2b(1+ctgb)} = sqrt{cosb(cosb+sinb) + sinb(sinb+cosb)} = sqrt{(sinb+cosb)(sinb+cosb)} = sinb+cosb = -sinb-cosb$