Question

тригонометрия. 63-66.

Answer 1

$63)\frac\pi4 < a < \frac\pi2\\ \sqrt{(ctg^2a-tg^2a)cos2a}*tg2a = \sqrt{\frac{cos^22a}{sin^2acos^2a}}*tg2a = \frac{cos2a}{sinacos}*\frac{2sinacosa}{cos2a} = 2\\64) \frac\pi2 < a < \frac34\pi = > sina>0, cosa, tga, ctg < 0\\ \sqrt{(ctga-tga)*2ctg2a} * tg2a +2 = \sqrt{\frac{cos2a}{sinacosa}*2\frac{cos2a}{sin2a}}}*tg2a + 2 = \sqrt{\frac{cos^22a}{sin^2acos^2a}}*tg2a + 2 = -\frac{cos2a}{sinacosa}*\frac{2sinacosa}{cos2a} + 2 = -2+2 = 0$

$65)\frac32\pi<a<2\pi => cosa > 0, sina, tga, ctga < 0\\\sqrt{sin^2a(1-ctga)+cos^2a(1-tga)} = \sqrt{sina(sina-cosa) - cosa(sina-cosa)} = \\ </p><p>=\sqrt{(sina-cosa)(sina-cosa)} = sina-cosa$

$66)\pi < b < \frac32\pi, sinb, cosb < 0, tgb, ctgb > 0\\\sqrt{cos^2b(1+tgb) + sin^2b(1+ctgb)} = \sqrt{cosb(cosb+sinb) + sinb(sinb+cosb)} = \sqrt{(sinb+cosb)(sinb+cosb)} = sinb+cosb = -sinb-cosb$