$\dfrac{2}{3}$ $sin^4 45° + sec^2 45 - 2 tan^2 60°$ is equal to

The correct answer is $\dfrac{-23}{6}$

Using values for angles,
$\dfrac{2}{3}$ $sin^4 45° + sec^2 45 - 2 tan^2 60°$
= $\dfrac{2}{3}$ ($\dfrac{√2}{2}$)⁴ + ($\dfrac{2}{√2}$)² - 2 (√3)²
= $\dfrac{2}{3}$ ($\dfrac{4}{16}$) + $\dfrac{4}{2}$ - 2 (3)
= $\dfrac{1}{6}$ × $\dfrac{2}{1}$ - 6
= $\dfrac{(13 - 26)}{6}$
= $\dfrac{- 23}{6}$