Question

The question is in the image. Answer the question 6.

Answer 1

Question 6.

Using the pythagorean theorem to find the value of r:

$r=\sqrt{x^2+y^2}=\sqrt{3^2+1^2}=\sqrt{9+1}=\sqrt{10}$

Next, using:

x = 3

y = 1

The six trigonometric functions are given by:

$\begin{gathered} \sin\theta=\frac{y}{r}=\frac{1}{\sqrt{10}} \\ \tan\theta=\frac{y}{x}=\frac{1}{3} \\ sec\theta=\frac{r}{x}=\frac{\sqrt{10}}{3} \\ \cos\theta=\frac{x}{r}=\frac{3}{\sqrt{10}} \\ csc\theta=\frac{r}{y}=\frac{\sqrt{10}}{1} \\ cot\theta=\frac{x}{y}=\frac{3}{1}=3 \end{gathered}$