Given the expression:
[tex]\frac{\tan(-\frac{2\pi}{3})}{\sin(\frac{7\pi}{4})}-\sec (-\pi)[/tex]Let's find the exact value of the expression.
• First rewrite tan(-2π/3) in terms of sines and cosines.
We have:
[tex]\frac{\frac{\sin(-\frac{2\pi}{3})}{\cos(-\frac{2\pi}{3})}}{\sin(\frac{7\pi}{4})}-\sec (-\pi)[/tex]• Now, rewrite in product form:
[tex]\begin{gathered} \frac{\sin(-\frac{2\pi}{3})}{\cos(-\frac{2\pi}{3})\times\sin(\frac{7\pi}{4})}-\sec (-\pi) \\ \\ \end{gathered}[/tex]Apply the reference angles by finding the angle suing the equivalent trigonomteric values.
We have:
[tex]\begin{gathered} \frac{\sin(\frac{4\pi}{3})}{\cos(\frac{4\pi}{3})(-\sin(\frac{\pi}{4})}-\sec (-\pi) \\ \\ \frac{-\sin(\frac{\pi}{3})}{-\cos(\frac{\pi}{3})(-\frac{\sin\pi}{4})}-\sec (-\pi) \end{gathered}[/tex]Substitute each trig value with the exact value:
[tex]\begin{gathered} \frac{-\frac{\sqrt[]{3}}{2}}{-\frac{1}{2}\times(-1\frac{\sqrt[]{2}}{2})}-\sec (-\pi) \\ \\ \frac{-\frac{\sqrt[]{3}}{2}}{-(\frac{-1}{2}\times\frac{\sqrt[]{2}}{2})_{}}-\sec (-\pi) \\ \\ \frac{-\frac{\sqrt[]{3}}{2}}{-(-\frac{\sqrt[]{2}}{4})}--\sec (0) \\ \\ \frac{-\sqrt[]{3}\times4}{2\times\sqrt[]{2}}--\sec (0) \\ \\ \frac{-2\sqrt[]{3}}{\sqrt[]{2}}--\sec (0) \\ \\ \end{gathered}[/tex]Rationalize the denominator:
[tex]\begin{gathered} \frac{-2\sqrt[]{3}}{\sqrt[]{2}}\times\frac{\sqrt[]{2}}{\sqrt[]{2}}+\sec (0) \\ \\ -\sqrt[]{3\times2}+\sec (0) \\ \\ -\sqrt[]{6}+\sec (0) \end{gathered}[/tex]Where:
sec(0) = 1
[tex]\begin{gathered} -\sqrt[]{6}+1 \\ \\ \end{gathered}[/tex]ANSWER:
[tex]\begin{gathered} \\ -\text{ }\sqrt[]{6}+1 \end{gathered}[/tex]