We need to evaluate the expression:
[tex]\frac{\tan(-\frac{2\pi}{3})}{\sin(\frac{7\pi}{4})}-\sec (-\pi)[/tex]
So, let's find the value of each trigonometric function separately, and then we can evaluate the whole expression.
We have:
[tex]\tan (-\frac{2\pi}{3})=\tan (\frac{\pi}{3})=\sqrt[]{3}[/tex]
Also:
[tex]\sin (\frac{7\pi}{4})=-\sin (\frac{\pi}{4})=-\frac{1}{\sqrt[]{2}}[/tex]
And:
[tex]\begin{gathered} \sec (-\pi)=\frac{1}{\cos(-\pi)}=\frac{1}{\cos \pi}=\frac{1}{-1} \\ \\ -\sec (-\pi)=1 \end{gathered}[/tex]
Therefore, the value of the whole expression is:
[tex]\begin{gathered} \frac{\sqrt[]{3}}{-\frac{1}{\sqrt[]{2}}}+1 \\ \\ \sqrt[]{3}\cdot(-\sqrt[]{2})+1 \\ \\ -\sqrt[]{6}+1 \\ \\ 1-\sqrt[]{6} \end{gathered}[/tex]
Thus, the answer is:
[tex]1-\sqrt[]{6}[/tex]
