Question

Let 0 be an angle in quadrant II such that

Answer 1

Solution:

Given:

$\sec \theta=-\frac{8}{7}$

Recall the inverse identity of trigonometry,

$\sec \theta=\frac{1}{\cos \theta}$

Hence,

$\begin{gathered} \sec \theta=-\frac{8}{7} \\ \frac{1}{\cos \theta}=-\frac{8}{7} \\ \text{Hence,} \\ \cos \theta=-\frac{7}{8} \end{gathered}$

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,

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

Also,

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

Therefore,

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