Given:
[tex]\frac{\tan(-\frac{2\pi}{3})}{\sin(\frac{7\pi}{4})}-\sec (-\pi)[/tex]
To find the exact value of the given trigonometric expression, we follow the process below:
We note first that sec(- π) is equal to:
[tex]\frac{1}{\text{cos}(-\pi)}[/tex]
So,
[tex]\frac{\tan(-\frac{2\pi}{3})}{\sin(\frac{7\pi}{4})}-\sec (-\pi)=\frac{\tan(-\frac{2\pi}{3})}{\sin(\frac{7\pi}{4})}-\frac{1}{\text{cos}(-\pi)}[/tex]
We also note that:
[tex]\sin (\frac{7\pi}{4})=-\frac{\sqrt[]{2}}{2}[/tex][tex]\text{cos}(-\pi)=-1[/tex]
Hence,
[tex]\frac{\tan(-\frac{2\pi}{3})}{\sin(\frac{7\pi}{4})}-\frac{1}{\text{cos}(-\pi)}=\frac{\tan(-\frac{2\pi}{3})}{-\frac{\sqrt[]{2}}{2}}-\frac{1}{-1}[/tex]
We can simplify this into:
[tex]\begin{gathered} \frac{\tan(-\frac{2\pi}{3})}{-\frac{\sqrt[]{2}}{2}}-\frac{1}{-1}=\frac{\tan(-\frac{2\pi}{3})-\frac{1}{\sqrt[]{2}}}{-\frac{1}{\sqrt[]{2}}} \\ \frac{\tan(-\frac{2\pi}{3})-\frac{1}{\sqrt[]{2}}}{-\frac{1}{\sqrt[]{2}}}=-\frac{(\tan(-\frac{2\pi}{3})-\frac{1}{\sqrt[]{2}})\sqrt[]{2}}{1} \\ \end{gathered}[/tex]
Simplify further into:
[tex]\begin{gathered} =-(\tan (-\frac{2\pi}{3})\sqrt[]{2}-1 \\ =-\sqrt[]{2}(\tan (-\frac{2\pi}{3})+1 \\ =\sqrt[]{2}(\tan (\frac{2\pi}{3})+1 \\ =\sqrt[]{2}(-\sqrt[]{3})+1 \\ =-\sqrt[]{6}+1 \end{gathered}[/tex]
Therefore, the answer is:
[tex]=-\sqrt[]{6}+1[/tex]
