Roger Smith

2021-12-28

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

3. ${\mathrm{sin}}^{2}\left(\theta \right)+{\mathrm{cos}}^{2}\left(\theta \right)=?$

Joseph Fair

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

$-\sqrt{{12}^{2}-{\left(-5\right)}^{2}}$
$=-\sqrt{144-25}$
$=-\sqrt{119}$
Step 2
1) Use $\mathrm{tan}\left(\theta \right)$= opposite / adjacent and $\mathrm{cot}\left(\theta \right)=adjacent/opposite$
$\mathrm{tan}\left(\theta \right)\mathrm{cot}\left(\theta \right)$
$=\left(\frac{-\sqrt{119}}{-5}\right)\left(\frac{-5}{-\sqrt{119}}\right)$
$=1$

scomparve5j

Step 3
2) Use $\mathrm{csc}\left(\theta \right)=hypotenuse/opposite$ and
$\mathrm{csc}\left(\theta \right)\mathrm{tan}\left(\theta \right)$
$=\left(\frac{12}{-\sqrt{119}}\right)\cdot \left(\frac{-\sqrt{119}}{-5}\right)$
$=-\frac{12}{5}$

karton

Step 4
3) From the triangle use

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