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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]\theta[/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 the problem, let's start by analyzing the given information and follow a step-by-step approach to find each of the trigonometric functions.

We are given:
[tex]\[ \cot \theta = -\frac{2}{17} \][/tex]
and it is stated that [tex]\(\theta\)[/tex] is in Quadrant IV.

Recall that:
[tex]\[ \cot \theta = \frac{\cos \theta}{\sin \theta} \][/tex]
Given [tex]\(\cot \theta\)[/tex]:
[tex]\[ \cot \theta = -\frac{2}{17} \implies \frac{\cos \theta}{\sin \theta} = -\frac{2}{17} \][/tex]

In Quadrant IV, [tex]\(\cos \theta\)[/tex] is positive and [tex]\(\sin \theta\)[/tex] is negative. Let’s assign the opposite side being 2 and the adjacent side being 17 (which satisfies the given [tex]\(\cot \theta\)[/tex]). We need to find the hypotenuse using the Pythagorean theorem:
[tex]\[ \text{hypotenuse} = \sqrt{2^2 + 17^2} = \sqrt{4 + 289} = \sqrt{293} \][/tex]

Now, we can find [tex]\(\sin \theta\)[/tex] and [tex]\(\cos \theta\)[/tex]:
[tex]\[ \sin \theta = -\frac{2}{\sqrt{293}} \quad (\text{negative in IV quadrant}) \][/tex]
[tex]\[ \cos \theta = \frac{17}{\sqrt{293}} \quad (\text{positive in IV quadrant}) \][/tex]

Next, let's find the remaining trigonometric functions:
[tex]\[ \csc \theta = \frac{1}{\sin \theta} = \frac{1}{-\frac{2}{\sqrt{293}}} = -\frac{\sqrt{293}}{2} \][/tex]
[tex]\[ \sec \theta = \frac{1}{\cos \theta} = \frac{1}{\frac{17}{\sqrt{293}}} = \frac{\sqrt{293}}{17} \][/tex]
[tex]\[ \tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{-\frac{2}{\sqrt{293}}}{\frac{17}{\sqrt{293}}} = -\frac{2}{17} \][/tex]

So, we have found the following values for the trigonometric functions of [tex]\(\theta\)[/tex]:
[tex]\[ \sin \theta = -\frac{2}{\sqrt{293}} \][/tex]
[tex]\[ \cos \theta = \frac{17}{\sqrt{293}} \][/tex]
[tex]\[ \csc \theta = -\frac{\sqrt{293}}{2} \][/tex]
[tex]\[ \sec \theta = \frac{\sqrt{293}}{17} \][/tex]
[tex]\[ \tan \theta = -\frac{2}{17} \][/tex]

The correct set of values among the given options is:
[tex]\[ \sin \theta = -\frac{2}{\sqrt{293}}, \cos \theta = \frac{17}{\sqrt{293}}, \csc \theta = -\frac{\sqrt{293}}{2}, \sec \theta = \frac{\sqrt{293}}{17}, \tan \theta = -\frac{2}{17} \][/tex]

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