Given:
[tex]\frac{\tan(-\frac{2\pi}{3})}{\sin(\frac{7\pi}{4})}-\sec (-\pi)[/tex]First
Recall
[tex]\sec (\pi)=\frac{1}{\cos (\pi)}[/tex]This implies
[tex]\sec (-\pi)=\frac{1}{\cos (-\pi)}[/tex]Hence, the given expression becomes
[tex]\frac{\tan(-\frac{2\pi}{3})}{\sin(\frac{7\pi}{4})}-\frac{1}{\cos (-\pi)}[/tex]Using calculator
[tex]\cos (-\pi)=-1[/tex]Hence,
[tex]\begin{gathered} \frac{\tan(-\frac{2\pi}{3})}{\sin(\frac{7\pi}{4})}-\frac{1}{\cos (-\pi)} \\ =\frac{\tan(-\frac{2\pi}{3})}{\sin(\frac{7\pi}{4})}-\frac{1}{-1} \\ =\frac{\tan(-\frac{2\pi}{3})}{\sin(\frac{7\pi}{4})}+1 \end{gathered}[/tex]Also,
[tex]\sin (\frac{7\pi}{4})=-\frac{1}{\sqrt[]{2}}[/tex]This gives
[tex]\begin{gathered} \frac{\tan(-\frac{2\pi}{3})}{\sin(\frac{7\pi}{4})}+1 \\ =\frac{\tan(-\frac{2\pi}{3})}{-\frac{1}{\sqrt[]{2}}}+1 \end{gathered}[/tex]Simplifying the expression gives
[tex]\begin{gathered} \frac{\tan(-\frac{2\pi}{3})}{-\frac{1}{\sqrt[]{2}}}+1 \\ =\tan (-\frac{2\pi}{3})\div-\frac{1}{\sqrt[]{2}}+1 \\ =\tan (-\frac{2\pi}{3})\times-\sqrt[]{2}+1 \\ =-\sqrt[]{2}\tan (-\frac{2\pi}{3})+1 \end{gathered}[/tex]Using the following property
[tex]\tan (-x)=-\tan (x)[/tex]It follows
[tex]\tan (-\frac{2\pi}{3})=-\tan (\frac{2\pi}{3})[/tex]Hence the expression becomes
[tex]\begin{gathered} -\sqrt[]{2}\tan (-\frac{2\pi}{3})+1 \\ =-\sqrt[]{2}(-\tan (\frac{2\pi}{3}))+1 \\ =\sqrt[]{2}\tan (\frac{2\pi}{3})+1 \end{gathered}[/tex]And
[tex]\tan (\frac{2\pi}{3})=-\sqrt[]{3}[/tex]This gives
[tex]\begin{gathered} \sqrt[]{2}\tan (-\frac{2\pi}{3})+1 \\ =\sqrt[]{2}(-\sqrt[]{3)}+1 \end{gathered}[/tex]Simplifying the expression gives
[tex]\begin{gathered} \sqrt[]{2}(-\sqrt[]{3)}+1 \\ =-\sqrt[]{6}+1 \end{gathered}[/tex]Therefore, the answer is
[tex]-\sqrt[]{6}+1[/tex]