Question

Really need some help, struggling to solve This practice from my ACT prep guide

Answer 1

Given:

`(\tan(-(2\pi)/(3)))/(\sin((7\pi)/(4)))-\sec (-\pi)`

To find the exact value of the given trigonometric expression, we follow the process below:

We note first that sec(- π) is equal to:

`\frac{1}{\text{cos}(-\pi)}`

So,

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

We also note that:

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

Hence,

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

We can simplify this into:

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

Simplify further into:

`\begin{gathered} =-(\tan (-(2\pi)/(3))\sqrt[]{2}-1 \\ =-\sqrt[]{2}(\tan (-(2\pi)/(3))+1 \\ =\sqrt[]{2}(\tan ((2\pi)/(3))+1 \\ =\sqrt[]{2}(-\sqrt[]{3})+1 \\ =-\sqrt[]{6}+1 \end{gathered}`

Therefore, the answer is:

`=-\sqrt[]{6}+1`