Question

если можно распишите понятнее​

Answer 1

Ответ:

1.

$tg( \frac{\pi}{3} ) + 2ctg( \frac{\pi}{6} ) = \frac{ \sin( \frac{\pi}{3} ) }{ \cos( \frac{\pi}{3} ) } + 2 \times \frac{ \cos( \frac{\pi}{6} ) }{ \sin( \frac{\pi}{6} ) } = \\ = \frac{ \frac{ \sqrt{3} }{2} }{ \frac{1}{2} } + 2 \times \frac{ \frac{ \sqrt{3} }{2} }{ \frac{1}{2} } = \frac{ \sqrt{3} }{2} \times 2 + 2 \times 2 \times \frac{ \sqrt{3} }{2} = \\ = \sqrt{3} + 2 \sqrt{3} = 3 \sqrt{3}$

3.

$3 \cos( \frac{\pi}{3} ) + 2 \sin( \frac{\pi}{4} ) = 3 \times \frac{1}{2} + 2 \times \frac{ \sqrt{2} }{2} = \\ = \frac{3}{2} + \sqrt{2} = \frac{3 + 2 \sqrt{2} }{2}$

2.

$tg( \frac{\pi}{4} ) - 3ctg( \frac{5\pi}{6} ) = 1 - 3 \times \frac{ \cos( \frac{5\pi}{6} ) }{ \sin( \frac{5\pi}{6} ) } = \\ = 1 - 3 \times \frac{ - \frac{ \sqrt{3} }{2} }{ \frac{1}{2} } = 1 + 3 \times \frac{ \sqrt{3} }{2} \times 2 = \\ = 1 + 3 \sqrt{3}$

4.

$- \sin( \frac{\pi}{2} ) + 2ctg( \frac{3\pi}{4} ) = - 1 - 2 = - 3$