Answer
$\displaystyle \frac{7}{2}$
Work Step by Step
reference angles:
$\left[\begin{array}{lllll}
Quadr.: & I & II & III & IV\\
\theta' & \theta & 180^{o}-\theta & \theta-180^{o} & 360^{o}-\theta
\end{array}\right]$
$300^{o}$ is in quadrant IV, where secant is positive,
reference angle: $360^{o}-300^{o}=60^{o}$
Function Values of Special Angles$: \sec 300^{o}=\sec 60^{o}=2$
$150^{o}$ is in quadrant II, where cosine is negative,
reference angle: $180^{o}-150^{o}=30^{o}$
Function Values of Special Angles:
$ \displaystyle \cos 240^{o}=-\cos 30^{o}=-\frac{\sqrt{3}}{2}$
$\tan 4 $is in quadrant I,
$\tan 45^{o}=1$
$\sec^{2}300^{o}-2\cos^{2}150^{o}+\tan 45^{o}=$
$=2^{2}-2(-\displaystyle \frac{\sqrt{3}}{2})^{2}+1$
$=4-2(\displaystyle \frac{3}{4})+1$
$=4-\displaystyle \frac{3}{2}+1$
$=\displaystyle \frac{7}{2}$
