Question

E = tan(120 degrees) + 3sen * (150 degrees) - cos(405 degrees)​

Answer 1

$\Large\boxed{\begin{array}{l}\sf E=tg(120^\circ)+3sen(150^\circ)-cos(405^\circ)\\\sf tg(120^\circ)=-tg(60^\circ)=-\sqrt{3}\\\sf sen(150^\circ)=sen(30^\circ)=\dfrac{1}{2}\\\sf cos(405^\circ)=cos(45^\circ)=\dfrac{\sqrt{2}}{2}\\\\\sf E=-\sqrt{3}+3\cdot\dfrac{1}{2}-\dfrac{\sqrt{2}}{2}\\\\\sf E=\dfrac{-2\sqrt{3}+3-\sqrt{2}}{2}\end{array}}$