If `tantheta =-4/3," then " sintheta` is

Options

A. `-4/5" but not "4/5`
B. `-4/5or4/5`
C. `4/5" but not "-4/5`
D. None of these

Explanation

To find the value of \( \sin \theta \) given that \( \tan \theta = -\frac{4}{3} \), we can follow these steps:

### Step 1: Identify the Quadrant
Since \( \tan \theta \) is negative, we know that \( \theta \) must be in either the second or fourth quadrant. In the second quadrant, \( \sin \theta \) is positive, while in the fourth quadrant, \( \sin \theta \) is negative.

**Hint:** Remember that \( \tan \theta = \frac{\sin \theta}{\cos \theta} \). The sign of \( \tan \theta \) tells you about the signs of \( \sin \theta \) and \( \cos \theta \).

### Step 2: Use the Definition of Tangent
We know that:
\[ \tan \theta = \frac{\sin \theta}{\cos \theta} \]
Given \( \tan \theta = -\frac{4}{3} \), we can set:
\[ \sin \theta = -4k \quad \text{and} \quad \cos \theta = 3k \]
for some positive value \( k \).

**Hint:** This step helps to express \( \sin \theta \) and \( \cos \theta \) in terms of a common variable.

### Step 3: Use the Pythagorean Identity
Using the identity \( \sin^2 \theta + \cos^2 \theta = 1 \):
\[ (-4k)^2 + (3k)^2 = 1 \]
This simplifies to:
\[ 16k^2 + 9k^2 = 1 \]
\[ 25k^2 = 1 \]
\[ k^2 = \frac{1}{25} \]
\[ k = \frac{1}{5} \]

**Hint:** The Pythagorean identity is crucial for relating sine and cosine values.

### Step 4: Find \( \sin \theta \) and \( \cos \theta \)
Now substituting \( k \) back into our expressions for \( \sin \theta \) and \( \cos \theta \):
\[ \sin \theta = -4k = -4 \times \frac{1}{5} = -\frac{4}{5} \]
\[ \cos \theta = 3k = 3 \times \frac{1}{5} = \frac{3}{5} \]

**Hint:** Substitute the value of \( k \) to find the specific values of sine and cosine.

### Step 5: Conclusion
Since we determined that \( \theta \) is in the second quadrant, where sine is positive, we take the positive value:
\[ \sin \theta = -\frac{4}{5} \]

Thus, the final answer is:
\[ \sin \theta = -\frac{4}{5} \]