Question

Let 0 be an angle in quadrant II such that

Answer 1

Given:

`\sec \theta=-(8)/(7)`

Recall the inverse identity of trigonometry,

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

Hence,

`\begin{gathered} \sec \theta=-(8)/(7) \\ (1)/(\cos \theta)=-(8)/(7) \\ \text{Hence,} \\ \cos \theta=-(7)/(8) \end{gathered}`

Using the identity of cosine,

`\begin{gathered} \cos \theta=\frac{\text{adjacent}}{hypotenuse} \\ \text{Thus, } \\ \text{adjacent}=-7 \\ \text{hypotenuse}=8 \end{gathered}`

Using the Pythagoras theorem to find the opposite,

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

`\begin{gathered} \text{hypotenuse}^2=opposite^2+adjacent^2 \\ 8^2=x^2+(-7^2) \\ 64=x^2+49 \\ 64-49=x^2 \\ 15=x^2 \\ x=\sqrt[]{15} \\ \text{opposite}=\sqrt[]{15} \end{gathered}`

Hence,

From the right triangle,

`\begin{gathered} \text{opposite}=\sqrt[]{15} \\ \text{adjacent}=-7 \\ \text{hypotenuse}=8 \end{gathered}`

Thus,

`\begin{gathered} \cot \theta \\ U\sin g\text{ the inverse trig identity;} \\ \cot \theta=(1)/(\tan\theta) \\ \text{Also, recall that } \\ \tan \theta=\frac{\text{opposite}}{adjacent} \\ \text{Hence,} \\ \cot \theta=(1)/(\tan\theta)=\frac{1}{\frac{\sqrt[]{15}}{-7}} \\ \cot \theta=-\frac{7}{\sqrt[]{15}} \end{gathered}`

Also,

`\begin{gathered} \sin \theta=\frac{\text{opposite}}{hypotenuse} \\ \sin \theta=\frac{\sqrt[]{15}}{8} \end{gathered}`

Therefore,