Question

тригонометрия
доказать равенство ​

Answer 1

$\displaystyle\bf\\tg30^\circ+tg40^\circ+tg50^\circ+tg60^\circ=\\\\\\=(tg30^\circ+tg60^\circ)+(tg40^\circ+tg50^\circ)=\\\\\\=\frac{\sqrt{3} }{3} +\sqrt{3} +\frac{Sin(40^\circ+50^\circ)}{Cos40^\circ\cdot Cos50^\circ} =\frac{4\sqrt{3} }{3} +\frac{Sin90}{Cos40^\circ\cdot Cos50^\circ}=\\\\\\=\frac{4\sqrt{3} }{3}+\frac{1}{Cos40^\circ Sin40^\circ }=\frac{4}{\sqrt3} }+\frac{2}{Cos10^\circ} =$

$\displaystyle\bf\\=\frac{4Cos10^\circ+2\sqrt{3} }{\sqrt{3} Cos10^\circ} } =\frac{4\Big(Cos10^\circ+\dfrac{\sqrt{3} }{2} \Big) }{\sqrt{3} Cos10^\circ} } =\\\\\\=\frac{4\Big(Cos10^\circ+Cos30^\circ \Big) }{\sqrt{3} Cos10^\circ} } =\frac{4\cdot 2 Cos20^\circ Cos10^\circ }{\sqrt{3} Cos10^\circ} } =\frac{8Cos20^\circ}{\sqrt{3} }$