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Use a Pythagorean identity to find [tex]\(\sec (\theta)\)[/tex] if [tex]\(\tan (\theta) = -9\)[/tex] and the terminal side of [tex]\(\theta\)[/tex] lies in quadrant II.

[tex]\(\sec (\theta) = \)[/tex] [tex]\(\square\)[/tex]

Sagot :

To find [tex]\(\sec (\theta)\)[/tex] given that [tex]\(\tan (\theta) = -9\)[/tex] and the terminal side of [tex]\(\theta\)[/tex] lies in quadrant II, follow these steps:

1. Understand Trigonometric Ratios in Quadrant II:
- In quadrant II, the tangent function ([tex]\(\tan(\theta) = \sin(\theta) / \cos(\theta)\)[/tex]) is negative because sine is positive and cosine is negative.

2. Express Tangent and Determine Sine and Cosine Components:
- Given [tex]\(\tan(\theta) = -9\)[/tex], we can write [tex]\(\sin(\theta) = 9k\)[/tex] and [tex]\(\cos(\theta) = -k\)[/tex] for some positive constant [tex]\(k\)[/tex], so:
[tex]\[ \frac{\sin(\theta)}{\cos(\theta)} = \frac{9k}{-k} = -9 \][/tex]

3. Use the Pythagorean Identity to Find the Hypotenuse ([tex]\(r\)[/tex]):
- The Pythagorean identity states that [tex]\(r^2 = \sin^2(\theta) + \cos^2(\theta)\)[/tex], where [tex]\(r\)[/tex] is the hypotenuse.
- Substitute [tex]\(\sin(\theta) = 9k\)[/tex] and [tex]\(\cos(\theta) = -k\)[/tex] into the identity:
[tex]\[ r = \sqrt{(9k)^2 + (-k)^2} = \sqrt{81k^2 + k^2} = \sqrt{82k^2} = \sqrt{82}k \][/tex]

4. Find [tex]\(\sec(\theta)\)[/tex] Using Its Definition:
- Secant function ([tex]\(\sec(\theta)\)[/tex]) is defined as:
[tex]\[ \sec(\theta) = \frac{1}{\cos(\theta)} \][/tex]
- Alternatively, considering the expression for the hypotenuse [tex]\(r\)[/tex] and cosine:
[tex]\[ \cos(\theta) = \frac{-k}{\sqrt{82}k} = \frac{-1}{\sqrt{82}} \][/tex]
- Therefore:
[tex]\[ \sec(\theta) = \frac{r}{\cos(\theta)} = \sqrt{82} \cdot \frac{1}{-1} = -\sqrt{82} \][/tex]
- If we simplify further:
[tex]\[ \sec(\theta) = -\sqrt{82} \][/tex]

Thus, the exact and fully simplified answer is:
[tex]\[ \sec(\theta) = -\sqrt{82} \][/tex]