Question

Помогите решить пожалуйста

Answer 1

$\frac{4}{3}\; tg (\pi -arcsin(-\frac{3}{5}) )= \frac{4}{3}\; tg( \pi +arcsin \frac{3}{5})= \frac{4}{3}\; tg(arcsin \frac{3}{5})=\\\\= \frac{4}{3} \cdot \frac{sin(arcsin\frac{3}{5})}{cos(arcsin\frac{3}{5})}= \frac{4}{3} \cdot \frac{\frac{3}{5}}{\sqrt{1-sin^2(arcsin\frac{3}{5})}} = \frac{4}{3} \cdot \frac{\frac{3}{5}}{\sqrt{1-(\frac{3}{5})^2}} =$

$= \frac{4}{3} \cdot \frac{3}{5\cdot \sqrt{1-\frac{9}{25}}} = \frac{4}{3} \cdot \frac{3}{5\cdot \sqrt{\frac{16}{25}}} = \frac{4}{3} \cdot \frac{3\cdot 5}{5\cdot 4}=1$

$\star \; \; sin(arcsin \alpha )= \alpha \; ,\; \; -1\leq \alpha \leq 1\; \; \; \star$