From the unit circle:
we can see that
[tex]\tan (-\frac{2\pi}{3})=\tan (\frac{4\pi}{3})=\frac{\frac{-\sqrt[]{3}}{2}}{-\frac{1}{2}}[/tex]
which gives
[tex]\tan (-\frac{2\pi}{3})=\tan (\frac{4\pi}{3})=\sqrt[]{3}[/tex]
Similarly,
[tex]\sin (\frac{7\pi}{4})=-\frac{\sqrt[]{2}}{2}[/tex]
and
[tex]\sec (-\pi)=\sec (\pi)=\frac{1}{\cos \pi}=\frac{1}{-1}=-1[/tex]
Therefore, by substituting ou last result into the given expression, we have
[tex]\frac{\tan (-\frac{2\pi}{3})}{\sin (\frac{7\pi}{4})}-\text{sec(-}\pi)=\frac{\sqrt[]{3}}{-\frac{\sqrt[]{2}}{2}}-(-1)[/tex]
then, we get
[tex]\frac{\tan(-\frac{2\pi}{3})}{\sin(\frac{7\pi}{4})}-\text{sec(-}\pi)=-\sqrt[]{2}\sqrt[]{3}-(-1)[/tex]
Therefore, the answer is:
[tex]\frac{\tan(-\frac{2\pi}{3})}{\sin(\frac{7\pi}{4})}-\text{sec(-}\pi)=-\sqrt[]{6}+1[/tex]
