Respuesta :
Given:
The point (9,-12) is on terminal side of angle theta in standard position.
To find:
The exact value of each of the six trigonometric functions of theta.
Solution:
The given point is (9,-12). Here, x-coordinate is positive and y-coordinate is negative. So, the point lies in 4th quadrant and only cos and sec are positive in 4th quadrant.
We know that,
[tex]r=\sqrt{x^2+y^2}[/tex]
[tex]r=\sqrt{9^2+(-12)^2}[/tex]
[tex]r=\sqrt{81+144}[/tex]
[tex]r=\sqrt{225}[/tex]
[tex]r=15[/tex]
Now,
[tex]\sin \theta=\dfrac{y}{r}=\dfrac{-12}{15}=-\dfrac{4}{5}[/tex]
[tex]\cos \theta=\dfrac{x}{r}=\dfrac{9}{15}=\dfrac{3}{5}[/tex]
[tex]\tan \theta=\dfrac{y}{x}=\dfrac{-12}{9}=-\dfrac{4}{3}[/tex]
[tex]\cot \theta=\dfrac{1}{\tan \theta}=-\dfrac{3}{4}[/tex]
[tex]\text{cosec} \theta=\dfrac{1}{\sin \theta}=-\dfrac{5}{4}[/tex]
[tex]\sec \theta=\dfrac{1}{\cos \theta}=\dfrac{5}{3}[/tex]
Therefore, the values of six trigonometric functions of theta are [tex]\sin \theta=-\dfrac{4}{5},\cos \theta=\dfrac{3}{5},\tan \theta=-\dfrac{4}{3},\cot \theta=-\dfrac{3}{4},\text{cosec} \theta=-\dfrac{5}{4},\sec \theta=\dfrac{5}{3}[/tex].
