Question

Буду рада ,если поможете;)

Answer 1

α - угол третьей четверти, значит tgα > 0 .

$1+tg^{2}\alpha=\frac{1}{Cos^{2}\alpha} \\\\tg\alpha ^{2}=\frac{1}{Cos^{2}\alpha}-1=\frac{1}{(-0,6)^{2} }-1=\frac{1}{0,36}-1=\frac{25}{9} -1=\frac{16}{9}\\\\tg\alpha=\sqrt{\frac{16}{9}}=\frac{4}{3}\\\\6tg\alpha=6*\frac{4}{3}=8\\\\\boxed{6tg\alpha=8}$

Второй способ :

$Sin\alpha=-\sqrt{1-Cos^{2}\alpha}=-\sqrt{1-(-0,6)^{2} }=-\sqrt{1-0,36}=-\sqrt{0,64}= -0,8\\\\tg\alpha=\frac{Sin\alpha }{Cos\alpha}=\frac{-0,8}{-0,6}=\frac{4}{3}\\\\6tg\alpha=6*\frac{4}{3} =8$