Question

Really need help solving It’s from my ACT prep guide

Answer 1

From the unit circle:

we can see that

`\tan (-(2\pi)/(3))=\tan ((4\pi)/(3))=\frac{\frac{-\sqrt[]{3}}{2}}{-(1)/(2)}`

which gives

`\tan (-(2\pi)/(3))=\tan ((4\pi)/(3))=\sqrt[]{3}`

Similarly,

`\sin ((7\pi)/(4))=-\frac{\sqrt[]{2}}{2}`

and

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

Therefore, by substituting ou last result into the given expression, we have

`(\tan (-(2\pi)/(3)))/(\sin ((7\pi)/(4)))-\text{sec(-}\pi)=\frac{\sqrt[]{3}}{-\frac{\sqrt[]{2}}{2}}-(-1)`

then, we get

`(\tan(-(2\pi)/(3)))/(\sin((7\pi)/(4)))-\text{sec(-}\pi)=-\sqrt[]{2}\sqrt[]{3}-(-1)`

Therefore, the answer is:

`(\tan(-(2\pi)/(3)))/(\sin((7\pi)/(4)))-\text{sec(-}\pi)=-\sqrt[]{6}+1`