Question

Даю 50 баллов. Решите номер 636 (в) с объяснением.​

Answer 1

Ответ:

$b)\ a=-\dfrac{3\pi}{4}\\\\tga+\dfrac{cosa}{sina-1}=\dfrac{sina}{cosa}+\dfrac{cosa}{sina-1}=\dfrac{sin^2a-sina+cos^2a}{cosa\, (sina-1)}=\dfrac{-sina}{cosa\, (sina-1)}=\\\\\\=\dfrac{-sin(-\dfrac{3\pi}{4})}{cos(-\dfrac{3\pi}{4})\cdot (sin(-\dfrac{3\pi}{4})-1)}=\dfrac{sin\dfrac{3\pi}{4}}{cos\dfrac{3\pi}{4}\cdot (-sin\dfrac{3\pi}{4}-1)}=\dfrac{\dfrac{\sqrt2}{2}}{-\dfrac{\sqrt2}{2}\cdot (-\dfrac{\sqrt2}{2}-1)}=\\\\\\=\dfrac{1}{\dfrac{\sqrt2+2}{2}}=\dfrac{2}{\sqrt2+2}=\dfrac{2(\sqrt2-2)}{2-4}=-(\sqrt2-2)=2-\sqrt2$

$c)\ \ a=\dfrac{7\pi}{3}\\\\ctg^2a\cdot (cos^2a-1)+2cos^2a=-\dfrac{cos^2a}{sin^2a}\cdot (1-cos^2a)+2cos^2a=\\\\\\=-\dfrac{cos^2a}{sin^2a}\cdot sin^2a+2cos^2a=-cos^2a+2cos^2a=cos^2a=cos^2\dfrac{7\pi}{3}=\\\\\\=cos^2(2\pi +\dfrac{\pi}{3})=cos^2\dfrac{\pi}{3}=\Big(\dfrac{1}{2}\Big)^2=\dfrac{1}{4}$