Question

Blake is told that an angle has a cosine of −
6
​

1
​
and is in Quadrant III. Use the Pythagorean identity sin
2
(θ)+cos
2
(θ)=1, the trigonometric identity tanθ=
cosθ
sinθ
​
, and the quadrant information to find the tangent.
□

Answer 1

It seems there may be a typo with the value of the cosine you provided. The cosine of an angle must be in the range of -1 to 1.

Assuming you meant the cosine of the angle, \(\cos(\theta) = -\frac{6}{1}=-6\), which is not valid, or perhaps you meant \( \cos(\theta) = -\frac{6}{\sqrt{1}} \) which also does not make sense as the resulting value is still outside of the valid range for cosine.

Instead, I will guide you on how to find the tangent of an angle in Quadrant III given a valid cosine value, for example, let's say \(\cos(\theta) = -\frac{3}{5}\).

### Step 1: Use the Pythagorean Identity

The Pythagorean identity states:
\[
\sin^2(\theta) + \cos^2(\theta) = 1
\]

Substituting our cosine value:
\[
\sin^2(\theta) + \left(-\frac{3}{5}\right)^2 = 1
\]

Calculating \(\cos^2(\theta)\):
\[
\sin^2(\theta) + \frac{9}{25} = 1
\]

Subtracting \(\frac{9}{25}\) from both sides:
\[
\sin^2(\theta) = 1 - \frac{9}{25} = \frac{25}{25} - \frac{9}{25} = \frac{16}{25}
\]

Finding \(\sin(\theta)\):
\[
\sin(\theta) = \pm \sqrt{\frac{16}{25}} = \pm \frac{4}{5}
\]

### Step 2: Determine the Sign of \(\sin(\theta)\)

Since the angle is in Quadrant III, both sine and cosine are negative:
\[
\sin(\theta) = -\frac{4}{5}
\]

### Step 3: Calculate \(\tan(\theta)\)

Using the identity for tangent:
\[
\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}
\]
Substituting our values:
\[
\tan(\theta) = \frac{-\frac{4}{5}}{-\frac{3}{5}} = \frac{-4}{-3} = \frac{4}{3}
\]

### Conclusion

So, the value of \(\tan(\theta)\) is \(\frac{4}{3}\).

Note: If there was a specific value for the cosine given that needed to be addressed, please clarify that value for further calculation.