Solved: "If cos⁡(θ)=−512 and θ is in quadrant III,..."

High School
Other

Roger Smith

Answered question

2021-12-28

If $\cos (θ) = - \frac{5}{12}$ and $θ$ is in quadrant III, then evaluate the following.
1. $\tan (θ) \cot (θ) = ?$
2. $\csc (θ) \tan (θ) = ?$

3. $\sin^{2} (θ) + \cos^{2} (θ) = ?$

Answer & Explanation

Joseph Fair

Beginner2021-12-29Added 34 answers

Step 1
Given that theta is in quadrant III , so we draw a triangle in quadrant III with adjacent $= - 5$ and hypotenuse $= 12$ .
Opposite = $- \sqrt{119}$
Negative because of the quadrant III

$- \sqrt{12^{2} - {(- 5)}^{2}}$
$= - \sqrt{144 - 25}$
$= - \sqrt{119}$
Step 2
1) Use $\tan (θ)$ = opposite / adjacent and $\cot (θ) = a d j a c e n t / o p p o s i t e$
$\tan (θ) \cot (θ)$
$= (\frac{- \sqrt{119}}{- 5}) (\frac{- 5}{- \sqrt{119}})$
$= 1$

scomparve5j

Beginner2021-12-30Added 38 answers

Step 3
2) Use $\csc (θ) = h y p o t e n u s e / o p p o s i t e$ and $\tan (θ) = o p p o s i t e / a d j a c e n t$
$\csc (θ) \tan (θ)$
$= (\frac{12}{- \sqrt{119}}) \cdot (\frac{- \sqrt{119}}{- 5})$
$= - \frac{12}{5}$

karton

Expert2022-01-04Added 613 answers

Step 4
3) From the triangle use

$\begin{matrix} \sin (θ) = - \sqrt{119} / 12 a n d \cos (θ) = - 5 / 12 \\ \sin^{2} (θ) + \cos^{2} (θ) \\ = (- \frac{\sqrt{119}}{12})^{2} (- \frac{5}{12})^{2} \\ = \frac{119}{144} + \frac{25}{144} \\ = \frac{144}{144} \\ = 1 \end{matrix}$

Do you have a similar question?

Recalculate according to your conditions!

New Questions in Other

Didn't find what you were looking for?