Answer:
[tex]-\sqrt[]{6}+1[/tex]Explanation:
Given the below expression;
[tex]\frac{\tan(-\frac{2\pi}{3})}{\sin(\frac{7\pi}{4})}-\sec (-\pi)[/tex]Recall that;
[tex]\begin{gathered} \sec x=\frac{1}{\cos x} \\ \sin x=\cos (\frac{\pi}{2}-x) \end{gathered}[/tex]So we can rewrite the expression as;
[tex]\begin{gathered} \frac{\tan(-\frac{2\pi}{3})}{\cos(\pi-\frac{7\pi}{4})}-\frac{1}{\cos(-\pi)} \\ \frac{\tan(-\frac{2\pi}{3})}{\cos(-\frac{5\pi}{4})}-\frac{1}{\cos(-\pi)} \end{gathered}[/tex]Also, recall that;
[tex]\begin{gathered} \cos (-x)=\cos x \\ \tan (-x)=-\tan x \end{gathered}[/tex]So we'll have;
[tex]\frac{-\tan (\frac{2\pi}{3})}{\cos (\frac{5\pi}{4})}-\frac{1}{\cos (\pi)}[/tex]From the Unit circle, we have that;
[tex]\begin{gathered} \cos \pi=-1 \\ \cos (\frac{5\pi}{4})=\frac{-\sqrt[]{2}}{2} \\ \tan (\frac{2\pi}{3})=-\sqrt[]{3} \end{gathered}[/tex]Substituting the above values into the expression and simplifying, we'll have;
[tex]\begin{gathered} \frac{-(-\sqrt[]{3})}{\frac{-\sqrt[]{2}}{2}}-\frac{1}{-1}=\frac{\sqrt[]{3}}{\frac{-\sqrt[]{2}}{2}}+1=-\frac{2\sqrt[]{3}\sqrt[]{2}}{\sqrt[]{2}\cdot\sqrt[]{2}}+1 \\ =-\sqrt[]{6}+1 \end{gathered}[/tex]