Question

the terminal side of an angle θ in standard position passes through the given point. evaluate csc θ, sec θ, and cot θ.

1. p(8, 6)
2. p(-8, 6)
3. p(-8, -6)
4. p(8, -6)

Explanation:

Step1: Find the value of $r$

For a point $P(x,y)$ on the terminal - side of an angle $\theta$ in standard position, $r=\sqrt{x^{2}+y^{2}}$.
For $P(8,6)$:
$r=\sqrt{8^{2}+6^{2}}=\sqrt{64 + 36}=\sqrt{100}=10$

Step2: Calculate $\csc\theta$

We know that $\csc\theta=\frac{r}{y}$.
$\csc\theta=\frac{10}{6}=\frac{5}{3}$

Step3: Calculate $\sec\theta$

We know that $\sec\theta=\frac{r}{x}$.
$\sec\theta=\frac{10}{8}=\frac{5}{4}$

Step4: Calculate $\cot\theta$

We know that $\cot\theta=\frac{x}{y}$.
$\cot\theta=\frac{8}{6}=\frac{4}{3}$

For $P(-8,6)$:
$r=\sqrt{(-8)^{2}+6^{2}}=\sqrt{64 + 36}=\sqrt{100}=10$
$\csc\theta=\frac{r}{y}=\frac{10}{6}=\frac{5}{3}$
$\sec\theta=\frac{r}{x}=\frac{10}{-8}=-\frac{5}{4}$
$\cot\theta=\frac{x}{y}=\frac{-8}{6}=-\frac{4}{3}$

For $P(-8,-6)$:
$r=\sqrt{(-8)^{2}+(-6)^{2}}=\sqrt{64 + 36}=\sqrt{100}=10$
$\csc\theta=\frac{r}{y}=\frac{10}{-6}=-\frac{5}{3}$
$\sec\theta=\frac{r}{x}=\frac{10}{-8}=-\frac{5}{4}$
$\cot\theta=\frac{x}{y}=\frac{-8}{-6}=\frac{4}{3}$

For $P(8,-6)$:
$r=\sqrt{8^{2}+(-6)^{2}}=\sqrt{64 + 36}=\sqrt{100}=10$
$\csc\theta=\frac{r}{y}=\frac{10}{-6}=-\frac{5}{3}$
$\sec\theta=\frac{r}{x}=\frac{10}{8}=\frac{5}{4}$
$\cot\theta=\frac{x}{y}=\frac{8}{-6}=-\frac{4}{3}$

1.

• $\csc\theta=\frac{5}{3}$, $\sec\theta=\frac{5}{4}$, $\cot\theta=\frac{4}{3}$

2.

• $\csc\theta=\frac{5}{3}$, $\sec\theta=-\frac{5}{4}$, $\cot\theta=-\frac{4}{3}$

3.

• $\csc\theta=-\frac{5}{3}$, $\sec\theta=-\frac{5}{4}$, $\cot\theta=\frac{4}{3}$

4.

• $\csc\theta=-\frac{5}{3}$, $\sec\theta=\frac{5}{4}$, $\cot\theta=-\frac{4}{3}$

Answer:

Step1: Find the value of $r$

For a point $P(x,y)$ on the terminal - side of an angle $\theta$ in standard position, $r=\sqrt{x^{2}+y^{2}}$.
For $P(8,6)$:
$r=\sqrt{8^{2}+6^{2}}=\sqrt{64 + 36}=\sqrt{100}=10$

Step2: Calculate $\csc\theta$

We know that $\csc\theta=\frac{r}{y}$.
$\csc\theta=\frac{10}{6}=\frac{5}{3}$

Step3: Calculate $\sec\theta$

We know that $\sec\theta=\frac{r}{x}$.
$\sec\theta=\frac{10}{8}=\frac{5}{4}$

Step4: Calculate $\cot\theta$

We know that $\cot\theta=\frac{x}{y}$.
$\cot\theta=\frac{8}{6}=\frac{4}{3}$

For $P(-8,6)$:
$r=\sqrt{(-8)^{2}+6^{2}}=\sqrt{64 + 36}=\sqrt{100}=10$
$\csc\theta=\frac{r}{y}=\frac{10}{6}=\frac{5}{3}$
$\sec\theta=\frac{r}{x}=\frac{10}{-8}=-\frac{5}{4}$
$\cot\theta=\frac{x}{y}=\frac{-8}{6}=-\frac{4}{3}$

For $P(-8,-6)$:
$r=\sqrt{(-8)^{2}+(-6)^{2}}=\sqrt{64 + 36}=\sqrt{100}=10$
$\csc\theta=\frac{r}{y}=\frac{10}{-6}=-\frac{5}{3}$
$\sec\theta=\frac{r}{x}=\frac{10}{-8}=-\frac{5}{4}$
$\cot\theta=\frac{x}{y}=\frac{-8}{-6}=\frac{4}{3}$

For $P(8,-6)$:
$r=\sqrt{8^{2}+(-6)^{2}}=\sqrt{64 + 36}=\sqrt{100}=10$
$\csc\theta=\frac{r}{y}=\frac{10}{-6}=-\frac{5}{3}$
$\sec\theta=\frac{r}{x}=\frac{10}{8}=\frac{5}{4}$
$\cot\theta=\frac{x}{y}=\frac{8}{-6}=-\frac{4}{3}$

1.

• $\csc\theta=\frac{5}{3}$, $\sec\theta=\frac{5}{4}$, $\cot\theta=\frac{4}{3}$

2.

• $\csc\theta=\frac{5}{3}$, $\sec\theta=-\frac{5}{4}$, $\cot\theta=-\frac{4}{3}$

3.

• $\csc\theta=-\frac{5}{3}$, $\sec\theta=-\frac{5}{4}$, $\cot\theta=\frac{4}{3}$

4.

• $\csc\theta=-\frac{5}{3}$, $\sec\theta=\frac{5}{4}$, $\cot\theta=-\frac{4}{3}$