step 1
Find out the value of
[tex]\tan (-\frac{2\pi}{3})[/tex]
Remember that
2pi/3=2(180)/3=120 degrees
the angle belonging to the third quadrant
the tangent in III quadrant is positive
so
tan(-2pi/3)=tan(180-120)=tan(60)
and
[tex]\begin{gathered} \tan (60^o)=\sqrt[]{3} \\ \tan (-\frac{2\pi}{3})=\sqrt[]{3} \end{gathered}[/tex]
step 2
Find out the value of
[tex]\sin (\frac{7\pi}{4})[/tex]
we have that
7pi/4=7*180/4=315 degrees
the angle lies on the IV quadrant
the sine is negative
sin(7pi/4)=sin (315)=-sin (360-315)=-sin(45)
[tex]\begin{gathered} \sin (45^o)=\frac{\sqrt[]{2}}{2} \\ \sin \text{ (}\frac{7\pi}{4})=-\frac{\sqrt[]{2}}{2} \end{gathered}[/tex]
step 3
Find out the value of
[tex]\sec (-\pi)[/tex]
sec(-pi)=sec(pi)=1/cos(pi)=1/-1=-1
step 4
substitute the given values in the original expression
[tex]\frac{\sqrt[]{3}}{\frac{-\sqrt[]{2}}{2}}-(-1)[/tex][tex]-\frac{2\sqrt[]{3}}{\sqrt[]{2}}+1=\frac{-2\sqrt[]{3}+\sqrt[]{2}}{\sqrt[]{2}}[/tex]
simplify
[tex]\frac{-2\sqrt[]{3}+\sqrt[]{2}}{\sqrt[]{2}}\cdot\frac{\sqrt[]{2}}{\sqrt[]{2}}=\frac{-2\sqrt[]{6}+2}{2}=-\sqrt[]{6}+1[/tex]
