Area of a triangle whose vertices are `(a cos theta, b sin theta) , ( - a sin theta , b cos theta) " and " ( - a cos theta, - b sin theta) ` is

Options

A. `ab sin theta cos theta `
B. `a cos theta sin theta`
C. `1/2 ab `
D. `ab`

Answer

To find the area of the triangle with vertices at \( (a \cos \theta, b \sin \theta) \), \( (-a \sin \theta, b \cos \theta) \), and \( (-a \cos \theta, -b \sin \theta) \), we can use the formula for the area of a triangle given its vertices \((x_1, y_1)\), \((x_2, y_2)\), and \((x_3, y_3)\):

\[ \text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \]

### Step 1: Identify the vertices
Let:
- \( (x_1, y_1) = (a \cos \theta, b \sin \theta) \)
- \( (x_2, y_2) = (-a \sin \theta, b \cos \theta) \)
- \( (x_3, y_3) = (-a \cos \theta, -b \sin \theta) \)

### Step 2: Substitute the vertices into the area formula
Substituting the coordinates into the area formula:

\[ \text{Area} = \frac{1}{2} \left| a \cos \theta \left( b \cos \theta - (-b \sin \theta) \right) + (-a \sin \theta) \left( -b \sin \theta - b \sin \theta \right) + (-a \cos \theta) \left( b \sin \theta - b \cos \theta \right) \right| \]

### Step 3: Simplify the expression
Now, simplify each term in the determinant:

1. The first term:
\[ a \cos \theta (b \cos \theta + b \sin \theta) = a b \cos \theta (\cos \theta + \sin \theta) \]

2. The second term:
\[ -a \sin \theta (-b \sin \theta - b \cos \theta) = a b \sin \theta (\sin \theta + \cos \theta) \]

3. The third term:
\[ -a \cos \theta (b \sin \theta - b \cos \theta) = -a b \cos \theta (\sin \theta - \cos \theta) \]

Combining these gives:
\[ \text{Area} = \frac{1}{2} \left| a b \cos \theta (\cos \theta + \sin \theta) + a b \sin \theta (\sin \theta + \cos \theta) - a b \cos \theta (\sin \theta - \cos \theta) \right| \]

### Step 4: Factor out common terms
Factoring out \( ab \):
\[ = \frac{1}{2} ab \left| \cos \theta (\cos \theta + \sin \theta) + \sin \theta (\sin \theta + \cos \theta) - \cos \theta (\sin \theta - \cos \theta) \right| \]

### Step 5: Combine and simplify
Combine the terms:
\[ = \frac{1}{2} ab \left| \cos^2 \theta + \sin^2 \theta + 2 \sin \theta \cos \theta \right| \]

Using the identity \( \sin^2 \theta + \cos^2 \theta = 1 \):
\[ = \frac{1}{2} ab \left| 1 + 2 \sin \theta \cos \theta \right| \]

### Step 6: Final simplification
Recognizing \( 2 \sin \theta \cos \theta = \sin(2\theta) \):
\[ = \frac{1}{2} ab \left| 1 + \sin(2\theta) \right| \]

### Final Result
Thus, the area of the triangle is:
\[ \text{Area} = ab \]