Question

ciga? 659. Упростите выражение: cos(-a) cos(180" + a). sin(-a) sin (90° + a) sin(-a) ctg(-a) 6) cos (360°-a). tg(180° + a) a) B) tg(180' - a) sin(210* + a) cos(180°-a) te(210-a)' cos(270° + a) etg(270'-a) sin (180 - 0) 18(270' + a) r) Только в и г ​

Answer 1

в)

$\displaystyle \frac{tg(180-\alpha )*sin(270+\alpha )}{cos(180-\alpha )*tg(270-\alpha )} = \frac{tg (\pi -\alpha )*sin(\frac{3\pi }{2} +\alpha )}{cos(\pi -\alpha )*tg(\frac{3\pi }{2} -\alpha )} =\\=\frac{-tg\alpha *(-cos\alpha )}{(-cos\alpha )*ctg\alpha } =-tg^{2} \alpha$

г)

$\displaystyle \frac{cos(270+\alpha)*ctg(270-\alpha )}{sin(180-\alpha )*tg(270+\alpha)} = \frac{cos(\frac{3\pi }{2} +\alpha)*ctg(\frac{3\pi }{2} -\alpha )}{sin(\pi -\alpha )*tg(\frac{3\pi }{2} +\alpha)} =\frac{sin\alpha *tg\alpha }{sin\alpha *(-ctg\alpha )} =\\=-tg^{2} \alpha$