Question

ajuda pf ( de preferencia bem detalhado)

Answer 1

$1320\° = 3\cdot360\°+240\°\\1320\°\,congruente\,a\,240\°\\1320\°\equiv240\°$

$2220\°=6\cdot360\°+60\°\\ 2200\°\equiv60\°$

$53\pi\equiv\pi\equiv180\°$

$\sec1320\°=\sec240\°=\dfrac{1}{\sin240\°}=\dfrac{1}{-\cos60\°}=-\dfrac{1}{\frac{1}{2}}=-2$

$\cos\dfrac{53\pi}{3}=\cos\dfrac{180\°}{3}=\cos60\°=\dfrac{\sqrt{3}}{2}$

$\tan2220\°=\tan60\°=\dfrac{\sin60\°}{\cos60\°}=\dfrac{\frac{\sqrt{3}}{2}}{\frac{1}{2}}=\sqrt{3}$

Agora a expressão pode ser avaliada:

$\dfrac{\sec1320\°}{2}-2\cdot\cos\left(\dfrac{53\pi}{3}\right)+(\tan2220\°)^2=\\\\\\\dfrac{\sec240\°}{2}-2\cdot\cos\left(\dfrac{180\°}{3}\right)+(\tan60\°)^2=\\\\\dfrac{\frac{1}{\cos240\°}}{2}-2\cdot\cos\left(60\°\right)+(\sqrt{3})^2=\\\\\dfrac{2}{-\cos60\°}-2\cdot\dfrac{1}{2}+3=\\\\-\dfrac{2}{\frac{1}{2}}-1+3=\\\\-4+3=\\\\\bold{-1}$