Question

Sabiendo que
t
a
n
(
α
+
β
)
= TAN a+ tan B/1-TAN a *TAN b, ENTONCES TAN 105°=?

Answer 1

Para este ejercicio usaremos los ángulos notables de 45° y 60°

tan(105°) = tan(45°+60°)

$tan(45+60)= \frac{tan(45)+tan(60)}{1-tan(45)tan(60)} }\\\\{tan(45+60)= \frac{1+ \sqrt{3} }{1-(1)( \sqrt{3)} } }\\\\{tan(45+60)= \frac{1+ \sqrt{3} }{1- \sqrt{3} } * \frac{1+ \sqrt{3}}{1+ \sqrt{3} } }\\\\{tan(45+60)= \frac{1+2 \sqrt{3} +3}{(1)^2-( \sqrt{3})^2 } }\\\\{tan45+60)= \frac{4+2 \sqrt{3} }{1-3} }\\\\{tan(45+60)= \frac{2(2+ \sqrt{3}) }{-2} }\\\\{tan(45+60)=-(2+ \sqrt{3} )}\\\\{tan(45+60)=-2- \sqrt{3} }\\\\{salu2. !!:)}$