Question

I really need help solving this practice from my prep guide in trigonometryI don’t know why but I have a hard time solving it completely

Answer 1

Given:

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

First

Recall

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

This implies

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

Hence, the given expression becomes

`(\tan(-(2\pi)/(3)))/(\sin((7\pi)/(4)))-(1)/(\cos (-\pi))`

Using calculator

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

Hence,

`\begin{gathered} (\tan(-(2\pi)/(3)))/(\sin((7\pi)/(4)))-(1)/(\cos (-\pi)) \\ =(\tan(-(2\pi)/(3)))/(\sin((7\pi)/(4)))-(1)/(-1) \\ =(\tan(-(2\pi)/(3)))/(\sin((7\pi)/(4)))+1 \end{gathered}`

Also,

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

This gives

`\begin{gathered} (\tan(-(2\pi)/(3)))/(\sin((7\pi)/(4)))+1 \\ =\frac{\tan(-(2\pi)/(3))}{-\frac{1}{\sqrt[]{2}}}+1 \end{gathered}`

Simplifying the expression gives

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

Using the following property

`\tan (-x)=-\tan (x)`

It follows

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

Hence the expression becomes

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

And

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

This gives

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

Simplifying the expression gives

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

Therefore, the answer is

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