Question

докажите что sin130⁰+sin 110⁰/cos130⁰+cos110⁰= -√3 ​

Answer 1

Ответ:

$-\sqrt{3}=-\sqrt{3}$

Объяснение:

$\dfrac{\sin 130^{\circ}+\sin 110^{\circ}}{\cos 130^{\circ}+\cos 110^{\circ}}=-\sqrt{3} \ ;$

$\dfrac{2\sin\dfrac{130^{\circ}+110^{\circ}}{2}\cos\dfrac{130^{\circ}-110^{\circ}}{2}}{2\cos\dfrac{130^{\circ}+110^{\circ}}{2}\cos\dfrac{130^{\circ}-110^{\circ}}{2}}=-\sqrt{3} \ ;$

$\dfrac{\sin 120^{\circ}}{\cos 120^{\circ}}=-\sqrt{3} \ ;$

$tg 120^{\circ}=-\sqrt{3} \ ;$

$tg(180^{\circ}-60^{\circ})=-\sqrt{3} \ ;$

$tg(-60^{\circ})=-\sqrt{3} \ ;$

$-tg 60^{\circ}=-\sqrt{3} \ ;$

$-\sqrt{3}=-\sqrt{3} \ ;$