Answer :
To determine the values of trigonometric functions given that \(\sec \theta = \frac{5}{3}\) and the terminal point determined by \(\theta\) is in quadrant IV, we will follow these steps:
1. Find \(\cos \theta\):
[tex]\[ \sec \theta = \frac{1}{\cos \theta} \][/tex]
Since \(\sec \theta = \frac{5}{3}\), we can rearrange to find \(\cos \theta\):
[tex]\[ \cos \theta = \frac{1}{\sec \theta} = \frac{1}{\frac{5}{3}} = \frac{3}{5} \][/tex]
Thus, \(\cos \theta = \frac{3}{5}\).
2. Find the hypotenuse and adjacent side of the right triangle:
Recall that:
[tex]\[ \cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}} \][/tex]
From \(\cos \theta = \frac{3}{5}\), we can say that if the hypotenuse is 5 and the adjacent side is 3.
3. Use the Pythagorean theorem to find the opposite side:
In a right triangle:
[tex]\[ \text{opposite}^2 + \text{adjacent}^2 = \text{hypotenuse}^2 \][/tex]
Substituting the values:
[tex]\[ \text{opposite}^2 + 3^2 = 5^2 \][/tex]
[tex]\[ \text{opposite}^2 + 9 = 25 \][/tex]
[tex]\[ \text{opposite}^2 = 16 \][/tex]
[tex]\[ \text{opposite} = 4 \][/tex]
Since \(\theta\) is in quadrant IV, where the sine function is negative, we have:
[tex]\[ \text{opposite} = -4 \][/tex]
4. Find \(\sin \theta\):
[tex]\[ \sin \theta = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{-4}{5} = -\frac{4}{5} \][/tex]
5. Find \(\tan \theta\):
[tex]\[ \tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{-\frac{4}{5}}{\frac{3}{5}} = -\frac{4}{3} \][/tex]
6. Find \(\csc \theta\):
[tex]\[ \csc \theta = \frac{1}{\sin \theta} = \frac{1}{-\frac{4}{5}} = -\frac{5}{4} \][/tex]
To summarize, the values of the trigonometric functions are:
- \(\cos \theta = \frac{3}{5}\)
- \(\sin \theta = -\frac{4}{5}\)
- \(\tan \theta = -\frac{4}{3}\)
- \(\csc \theta = -\frac{5}{4}\)
Now, let's check the given options:
A. \(\tan \theta = \frac{4}{3}\) is incorrect; \(\tan \theta = -\frac{4}{3}\).
B. \(\cos \theta = \frac{3}{5}\) is correct.
C. \(\sin \theta = -\frac{2}{5}\) is incorrect; \(\sin \theta = -\frac{4}{5}\).
D. \(\csc \theta = -\frac{5}{4}\) is correct.
Thus, the correct options are B and D.
1. Find \(\cos \theta\):
[tex]\[ \sec \theta = \frac{1}{\cos \theta} \][/tex]
Since \(\sec \theta = \frac{5}{3}\), we can rearrange to find \(\cos \theta\):
[tex]\[ \cos \theta = \frac{1}{\sec \theta} = \frac{1}{\frac{5}{3}} = \frac{3}{5} \][/tex]
Thus, \(\cos \theta = \frac{3}{5}\).
2. Find the hypotenuse and adjacent side of the right triangle:
Recall that:
[tex]\[ \cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}} \][/tex]
From \(\cos \theta = \frac{3}{5}\), we can say that if the hypotenuse is 5 and the adjacent side is 3.
3. Use the Pythagorean theorem to find the opposite side:
In a right triangle:
[tex]\[ \text{opposite}^2 + \text{adjacent}^2 = \text{hypotenuse}^2 \][/tex]
Substituting the values:
[tex]\[ \text{opposite}^2 + 3^2 = 5^2 \][/tex]
[tex]\[ \text{opposite}^2 + 9 = 25 \][/tex]
[tex]\[ \text{opposite}^2 = 16 \][/tex]
[tex]\[ \text{opposite} = 4 \][/tex]
Since \(\theta\) is in quadrant IV, where the sine function is negative, we have:
[tex]\[ \text{opposite} = -4 \][/tex]
4. Find \(\sin \theta\):
[tex]\[ \sin \theta = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{-4}{5} = -\frac{4}{5} \][/tex]
5. Find \(\tan \theta\):
[tex]\[ \tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{-\frac{4}{5}}{\frac{3}{5}} = -\frac{4}{3} \][/tex]
6. Find \(\csc \theta\):
[tex]\[ \csc \theta = \frac{1}{\sin \theta} = \frac{1}{-\frac{4}{5}} = -\frac{5}{4} \][/tex]
To summarize, the values of the trigonometric functions are:
- \(\cos \theta = \frac{3}{5}\)
- \(\sin \theta = -\frac{4}{5}\)
- \(\tan \theta = -\frac{4}{3}\)
- \(\csc \theta = -\frac{5}{4}\)
Now, let's check the given options:
A. \(\tan \theta = \frac{4}{3}\) is incorrect; \(\tan \theta = -\frac{4}{3}\).
B. \(\cos \theta = \frac{3}{5}\) is correct.
C. \(\sin \theta = -\frac{2}{5}\) is incorrect; \(\sin \theta = -\frac{4}{5}\).
D. \(\csc \theta = -\frac{5}{4}\) is correct.
Thus, the correct options are B and D.
