Question

Please Help Me

Answer 1

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,

$r=\sqrt{x^2+y^2}$

$r=\sqrt{9^2+(-12)^2}$

$r=\sqrt{81+144}$

$r=\sqrt{225}$

$r=15$

Now,

$\sin \theta=\dfrac{y}{r}=\dfrac{-12}{15}=-\dfrac{4}{5}$

$\cos \theta=\dfrac{x}{r}=\dfrac{9}{15}=\dfrac{3}{5}$

$\tan \theta=\dfrac{y}{x}=\dfrac{-12}{9}=-\dfrac{4}{3}$

$\cot \theta=\dfrac{1}{\tan \theta}=-\dfrac{3}{4}$

$\text{cosec} \theta=\dfrac{1}{\sin \theta}=-\dfrac{5}{4}$

$\sec \theta=\dfrac{1}{\cos \theta}=\dfrac{5}{3}$

Therefore, the values of six trigonometric functions of theta are $\sin \theta=-\dfrac{4}{5},\cos \theta=\dfrac{3}{5},\tan \theta=-\dfrac{4}{3},\cot \theta=-\dfrac{3}{4},\text{cosec} \theta=-\dfrac{5}{4},\sec \theta=\dfrac{5}{3}$.