If tan `x= -(3)/(4)` and x is in the second quadrant, then what is the value of `sinx*cosx`?

Options

A. `(6)/(25)`
B. `(12)/(25)`
C. `-(6)/(25)`
D. `-(12)/(25)`

Answer

To solve the problem where \( \tan x = -\frac{3}{4} \) and \( x \) is in the second quadrant, we need to find the value of \( \sin x \cdot \cos x \). Here’s the step-by-step solution:

### Step 1: Understand the given information
We know that:
- \( \tan x = -\frac{3}{4} \)
- \( x \) is in the second quadrant.

### Step 2: Set up a right triangle
In the second quadrant, the sine is positive and the cosine is negative. We can represent the tangent as the ratio of the opposite side (perpendicular) to the adjacent side (base) of a right triangle.

Let:
- Opposite side (perpendicular) = 3k
- Adjacent side (base) = -4k (negative because cosine is negative in the second quadrant)

Thus, we have:
\[ \tan x = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{3k}{-4k} = -\frac{3}{4} \]

### Step 3: Find the hypotenuse using the Pythagorean theorem
Using the Pythagorean theorem:
\[ h^2 = (3k)^2 + (-4k)^2 \]
\[ h^2 = 9k^2 + 16k^2 = 25k^2 \]
\[ h = \sqrt{25k^2} = 5k \]

### Step 4: Calculate \( \sin x \) and \( \cos x \)
Now we can find \( \sin x \) and \( \cos x \):
- \( \sin x = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{3k}{5k} = \frac{3}{5} \)
- \( \cos x = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{-4k}{5k} = -\frac{4}{5} \)

### Step 5: Calculate \( \sin x \cdot \cos x \)
Now we can find \( \sin x \cdot \cos x \):
\[ \sin x \cdot \cos x = \left(\frac{3}{5}\right) \cdot \left(-\frac{4}{5}\right) = -\frac{12}{25} \]

### Final Answer
Thus, the value of \( \sin x \cdot \cos x \) is:
\[ \boxed{-\frac{12}{25}} \]