Question

Hi, can you help me to solve this exercise, please!!

Answer 1

Given that

`\cot (\theta)=-\frac{\sqrt[]{35}}{7}`

Note:

`\cot (\theta)=(1)/(\tan (\theta))`

Therefore,

`(1)/(\tan (\theta))=-\frac{\sqrt[]{35}}{7}`

Cross-multiply

`\begin{gathered} 7*1=-\sqrt[]{35}*\tan (\theta) \\ 7=-\sqrt[]{35}\tan (\theta) \\ -\frac{7}{\sqrt[]{35}}=\tan (\theta) \\ \therefore\tan (\theta)=-\frac{7}{\sqrt[]{35}} \end{gathered}`

Using trigonometry ratios,

`\tan (\theta)=\frac{opposite}{\text{adjacent}}`

Where,

`\begin{gathered} \text{opposite}=-7 \\ \text{adjacent}=\sqrt[]{35} \end{gathered}`

Let us now use Pythagoras theorem to obtain the hypotenuse

The formula for the Pythagoras theorem is,

`\begin{gathered} \text{hypotenuse}^2=opposite^2+adjacent^2 \\ \\ \end{gathered}`

Hence,

`\begin{gathered} \text{hypotenuse}^2=(-7)^2+(\sqrt[]{35})^2 \\ \text{hypotenuse}^2=49+35=84 \\ \text{hypotenuse}=\sqrt[]{84}=\sqrt[]{4*21}=\sqrt[]{4}*\sqrt[]{21}=2*\sqrt[]{21}=2\sqrt[]{21} \\ \therefore\text{hypotenuse=}2\sqrt[]{21} \end{gathered}`

We are to solve for csc(θ),

`\csc (\theta)=(1)/(\sin (\theta))`

Where,

`\sin (\theta)=\frac{opposite}{\text{hypotenuse}}`

Hence,

`\sin (\theta)=-\frac{7}{2\sqrt[]{21}}`

Therefore,

`\begin{gathered} \csc (\theta)=\frac{1}{-\frac{7}{2\sqrt[]{21}}} \\ \csc (\theta)=1/-\frac{7}{2\sqrt[]{21}} \\ \csc (\theta)=1*-\frac{2\sqrt[]{21}}{7}=-\frac{2\sqrt[]{21}}{7} \\ \therefore\csc (\theta)=-\frac{2\sqrt[]{21}}{7} \end{gathered}`

Hence, the answer is

`\csc (\theta)=-\frac{2\sqrt[]{21}}{7}`