Question

1) (-sin 30° - cos 30°)² / 3cos 45° sin 45° - 6tg 30° ctg60° ;

​

Answer 1

-2-√3

Объяснение:

$\frac{(-\sin30^0-\cos30^0)^2}{3\cos45^0\sin 45^0-6\tan30^0\cot60^0}=\frac{(-\frac{1}{2}-\frac{\sqrt{3}}{2})^2}{3\frac{1}{\sqrt{2}}\frac{1}{\sqrt{2}}-6\frac{1}{\sqrt{3}}\frac{1}{\sqrt{3}}}= \frac{\frac{1}{4}+\frac{\sqrt{3}}{2}+\frac{3}{4}}{3\frac{1}{2}-6\frac{1}{{3}}}=\frac{1+\frac{\sqrt{3}}{2}}{\frac{3}{2}-2}=\\=-2\left(1+\frac{\sqrt{3}}{2}}\right)=-2-\sqrt{3}$