Question

I need help with this practice problem It’s from my ACT PREP GUIDE

Answer 1

step 1

Find out the value of

`\tan (-(2\pi)/(3))`

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

`\begin{gathered} \tan (60^o)=\sqrt[]{3} \\ \tan (-(2\pi)/(3))=\sqrt[]{3} \end{gathered}`

step 2

Find out the value of

`\sin ((7\pi)/(4))`

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)

`\begin{gathered} \sin (45^o)=\frac{\sqrt[]{2}}{2} \\ \sin \text{ (}(7\pi)/(4))=-\frac{\sqrt[]{2}}{2} \end{gathered}`

step 3

Find out the value of

`\sec (-\pi)`

sec(-pi)=sec(pi)=1/cos(pi)=1/-1=-1

step 4

substitute the given values in the original expression

`\frac{\sqrt[]{3}}{\frac{-\sqrt[]{2}}{2}}-(-1)`
`-\frac{2\sqrt[]{3}}{\sqrt[]{2}}+1=\frac{-2\sqrt[]{3}+\sqrt[]{2}}{\sqrt[]{2}}`

simplify

`\frac{-2\sqrt[]{3}+\sqrt[]{2}}{\sqrt[]{2}}\cdot\frac{\sqrt[]{2}}{\sqrt[]{2}}=\frac{-2\sqrt[]{6}+2}{2}=-\sqrt[]{6}+1`