Question

calcule o valor de tg 80° + tg 280° + tg 100° / tg80°

Answer 1

$\displaystyle \sf \frac{tg(80^\circ)+tg(280^\circ)+tg(100^\circ)}{tg(80^\circ)} \\\\\\ Sabemos \ que \ se : \\\\ a+b = 180^\circ \to a = 180^\circ - b \to tg(a) = tg(180^\circ-b)\to \boxed{\sf tg(a) = -tg(b) } \\\\\\ a - b = 180^\circ \to a=180^\circ +b \to tg(a) = tg(180^\circ + b) \to \boxed{\sf tg(a) = tg(b) }\\\\\ ent{\~a}o : \\\\\ tg(280) = tg(180+100) = tg(100^\circ) \\\\ tg(100^\circ) = tg(180-80) = -tg(80^\circ) \\\\\ Da{\'i}} :$

$\displaystyle \sf \frac{tg(80^\circ)+\overbrace{\sf tg(280^\circ)}^{tg(100^\circ)}+tg(100^\circ)}{tg(80^\circ)} \\\\\\ \frac{tg(80^\circ)+tg(100^\circ)+tg(100^\circ)}{tg(80^\circ)} \\\\\\ \frac{tg(80^\circ)+2\cdot \overbrace{\sf tg(100^\circ)}^{-tg(80^\circ)}}{tg(80^\circ)} \\\\\\\ \frac{tg(80^\circ)-2tg(80^\circ)}{tg(80^\circ)} \\\\\\\ \frac{-tg(80^\circ)}{tg(80^\circ)} = - 1 \\\\\ Portanto : \\\\ \boxed{\sf \frac{tg(80^\circ)+tg(280^\circ)+tg(100^\circ)}{tg(80^\circ)} = - 1 \ } \checkmark$