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Math  /  Trigonometry

QuestionLet θ\theta be an angle such that sec⁡θ=53\sec \theta=\frac{5}{3} and cot⁡θ<0\cot \theta<0. Find the exact values of tan⁡θ\tan \theta and csc⁡θ\csc \theta.

Studdy Solution

STEP 1

What is this asking? Given that secant of theta is 5/3 and cotangent of theta is negative, we need to find the exact values of tangent and cosecant of theta. Watch out! Don't forget to consider the signs of the trigonometric functions in the correct quadrant!

STEP 2

1. Determine the quadrant.
2. Find cos⁡θ \cos \theta .
3. Find sin⁡θ \sin \theta .
4. Calculate tan⁡θ \tan \theta .
5. Calculate csc⁡θ \csc \theta .

STEP 3

We are given that sec⁡θ=53 \sec \theta = \frac{5}{3} .
Since sec⁡θ \sec \theta is **positive**, cos⁡θ \cos \theta is also **positive**.
Cosine is positive in the **first** and **fourth** quadrants.

STEP 4

We are also given that cot⁡θ<0 \cot \theta < 0 , meaning cot⁡θ \cot \theta is **negative**.
Cotangent is negative in the **second** and **fourth** quadrants.

STEP 5

Since cos⁡θ \cos \theta is positive and cot⁡θ \cot \theta is negative, θ \theta must be in the **fourth quadrant**.

STEP 6

We know that sec⁡θ=1cos⁡θ \sec \theta = \frac{1}{\cos \theta} .

STEP 7

Given sec⁡θ=53 \sec \theta = \frac{5}{3} , we have 53=1cos⁡θ \frac{5}{3} = \frac{1}{\cos \theta} .
Therefore, cos⁡θ=35 \cos \theta = \frac{3}{5} .

STEP 8

We can use the Pythagorean identity: sin⁡2θ+cos⁡2θ=1 \sin^2 \theta + \cos^2 \theta = 1 .

STEP 9

Substituting cos⁡θ=35 \cos \theta = \frac{3}{5} , we get sin⁡2θ+(35)2=1 \sin^2 \theta + \left(\frac{3}{5}\right)^2 = 1 .

STEP 10

This simplifies to sin⁡2θ+925=1 \sin^2 \theta + \frac{9}{25} = 1 , so sin⁡2θ=1−925=2525−925=1625 \sin^2 \theta = 1 - \frac{9}{25} = \frac{25}{25} - \frac{9}{25} = \frac{16}{25} .

STEP 11

Taking the square root of both sides, we get sin⁡θ=±45 \sin \theta = \pm \frac{4}{5} .
Since θ \theta is in the **fourth quadrant**, where sine is **negative**, we have sin⁡θ=−45 \sin \theta = -\frac{4}{5} .

STEP 12

We know that tan⁡θ=sin⁡θcos⁡θ \tan \theta = \frac{\sin \theta}{\cos \theta} .

STEP 13

Substituting the values we found, we have tan⁡θ=−4535=−45⋅53=−43 \tan \theta = \frac{-\frac{4}{5}}{\frac{3}{5}} = -\frac{4}{5} \cdot \frac{5}{3} = -\frac{4}{3} .

STEP 14

We know that csc⁡θ=1sin⁡θ \csc \theta = \frac{1}{\sin \theta} .

STEP 15

Substituting sin⁡θ=−45 \sin \theta = -\frac{4}{5} , we get csc⁡θ=1−45=−54 \csc \theta = \frac{1}{-\frac{4}{5}} = -\frac{5}{4} .

STEP 16

We found that tan⁡θ=−43 \tan \theta = -\frac{4}{3} and csc⁡θ=−54 \csc \theta = -\frac{5}{4} .

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