Answered

Suppose [tex]$\theta$[/tex] is an angle in the standard position whose terminal side is in Quadrant IV and [tex]$\cot \theta = -\frac{2}{17}$[/tex]. Find the exact values of the five remaining trigonometric functions of [tex][tex]$\theta$[/tex][/tex].

A. [tex]\sin \theta = -\frac{2}{\sqrt{293}}, \cos \theta = \frac{17}{\sqrt{293}}, \csc \theta = -\frac{\sqrt{293}}{2}, \sec \theta = \frac{\sqrt{293}}{17}, \tan \theta = -\frac{17}{2}[/tex]

B. [tex]\sin \theta = -\frac{17}{\sqrt{293}}, \cos \theta = \frac{2}{\sqrt{293}}, \csc \theta = -\frac{\sqrt{293}}{17}, \sec \theta = \frac{\sqrt{293}}{2}, \tan \theta = -\frac{17}{2}[/tex]

C. [tex]\sin \theta = -\frac{17}{\sqrt{293}}, \cos \theta = \frac{2}{\sqrt{293}}, \csc \theta = -\frac{\sqrt{293}}{2}, \sec \theta = \frac{\sqrt{293}}{17}, \tan \theta = -\frac{17}{2}[/tex]

D. [tex]\sin \theta = \frac{\sqrt{293}}{17}, \cos \theta = -\frac{\sqrt{293}}{2}, \csc \theta = \frac{17}{\sqrt{293}}, \sec \theta = -\frac{2}{\sqrt{293}}, \tan \theta = -\frac{2}{17}[/tex]



Answer :

GinnyAnswer
To solve this problem, we need to find the exact values of the six trigonometric functions for an angle [tex]\(\theta\)[/tex] in Quadrant IV, given that [tex]\(\cot \theta = -\frac{2}{17}\)[/tex].

1. Cosecant ([tex]\(\csc \theta\)[/tex]) and secant ([tex]\(\sec \theta\)[/tex]) require [tex]\(\sin \theta\)[/tex] and [tex]\(\cos \theta\)[/tex]:
- Recall that [tex]\(\cot \theta = \frac{\cos \theta}{\sin \theta}\)[/tex], and we are given [tex]\(\cot \theta = -\frac{2}{17}\)[/tex].
- Let [tex]\(\sin \theta = -y\)[/tex] and [tex]\(\cos \theta = x\)[/tex] since we know the sine is negative and cosine is positive in Quadrant IV. This gives us:
[tex]\[ \frac{x}{-y} = -\frac{2}{17} \implies x = \frac{2}{17}y \][/tex]

2. Using the Pythagorean identity:
- The Pythagorean identity tells us that:
[tex]\[ \sin^2 \theta + \cos^2 \theta = 1 \][/tex]
- Substituting the values we have:
[tex]\[ (-y)^2 + \left( \frac{2}{17} y \right)^2 = 1 \][/tex]
- Simplifying this:
[tex]\[ y^2 + \left( \frac{2}{17} y \right)^2 = 1 \implies y^2 + \frac{4}{289} y^2 = 1 \implies y^2 \left( 1 + \frac{4}{289} \right) = 1 \][/tex]
- Further simplifying:
[tex]\[ y^2 \left( \frac{293}{289} \right) = 1 \implies y^2 = \frac{289}{293} \implies y = \frac{17}{\sqrt{293}} \][/tex]

3. Calculating [tex]\(\sin \theta\)[/tex] and [tex]\(\cos \theta\)[/tex]:
- Therefore, given that [tex]\( \sin \theta = -y \)[/tex]:
[tex]\[ \sin \theta = -\frac{17}{\sqrt{293}} \][/tex]
- And for [tex]\(\cos \theta\)[/tex]:
[tex]\[ \cos \theta = \frac{2}{17} y = \frac{2}{17} \times \frac{17}{\sqrt{293}} = \frac{2}{\sqrt{293}} \][/tex]

4. Finding the exact values for the other trigonometric functions:
- The cosecant ([tex]\(\csc \theta\)[/tex]):
[tex]\[ \csc \theta = \frac{1}{\sin \theta} = \frac{1}{-\frac{17}{\sqrt{293}}} = -\frac{\sqrt{293}}{17} \][/tex]
- The secant ([tex]\(\sec \theta\)[/tex]):
[tex]\[ \sec \theta = \frac{1}{\cos \theta} = \frac{1}{\frac{2}{\sqrt{293}}} = \frac{\sqrt{293}}{2} \][/tex]
- The tangent ([tex]\(\tan \theta\)[/tex]):
[tex]\[ \tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{-\frac{17}{\sqrt{293}}}{\frac{2}{\sqrt{293}}} = -\frac{17}{2} \][/tex]

Thus, the exact values of the five remaining trigonometric functions of [tex]\(\theta\)[/tex] are as follows:
[tex]\[ \sin \theta = -\frac{17}{\sqrt{293}}, \quad \cos \theta = \frac{2}{\sqrt{293}}, \quad \csc \theta = -\frac{\sqrt{293}}{17}, \quad \sec \theta = \frac{\sqrt{293}}{2}, \quad \tan \theta = -\frac{17}{2} \][/tex]

Hence, the correct answer from the given options is:
[tex]\[ \sin \theta=-\frac{17}{\sqrt{293}}, \cos \theta=\frac{2}{\sqrt{293}}, \csc \theta=-\frac{\sqrt{293}}{17}, \sec \theta=\frac{\sqrt{293}}{2}, \tan \theta=-\frac{17}{2} \][/tex]