Question

Срочно, 10 класс алгебра !!!!

Answer 1

Ответ:

$1. 1)\frac{\pi}{2} < \alpha < \pi;\\cos \alpha < 0;\\cos \alpha = -\sqrt{1-sin^2\alpha} = -\sqrt{\frac{16}{25}} = -\frac{4}{5};\\2) tg\alpha+1 = \frac{sin\alpha}{cos\alpha}+1 = -\frac{3}{4}+1 = \frac{1}{4}\\\\2. \alpha = \frac{\pi}{2};\\tg\frac{\alpha}{2}+sin(-\frac{\alpha}{3}) = tg\frac{\pi}{4}+sin(-\frac{\pi}{6}) = \\= 1+(-\frac{1}{2}) = \frac{1}{2};\\\\3. sin\alpha*cos\alpha(tg\alpha-ctg(-\alpha)) = \\= sin\alpha*cos\alpha(tg\alpha+ctg\alpha) = \\= cos^2\alpha+sin^2\alpha = 1.$