Question

Докажите тождество
2sin(pi/6-x)/cosx=1-sqrt(3)tgx

Answer 1

Доказательство:

$\dfrac{2sin\Big (\dfrac{\pi}{6}-x \Big ) }{cos~x} = 1 -\sqrt{3}\cdot tg~x$

$\dfrac{2\cdot \Big (sin\dfrac{\pi}{6}\cdot cos~x - sin~x\cdot cos\dfrac{\pi}{6} \Big ) }{cos~x} = 1 -\sqrt{3}\cdot tg~x$

$\dfrac{2\cdot (0.5\cdot cos~x - sin~x\cdot 0.5\sqrt{3} ) }{cos~x} = 1 -\sqrt{3}\cdot tg~x$

$\dfrac{ cos~x - \sqrt{3}\cdot sin~x }{cos~x} = 1 -\sqrt{3}\cdot tg~x$

$1 -\sqrt{3}\cdot tg~x \equiv 1 -\sqrt{3}\cdot tg~x$