If `alpha+beta=90^@` and `alpha=2beta`, then the value of `3cos^2alpha-2sin^2beta` is equal to:

Options

A. `3/4`
B. `1/4`
C. `4/3`
D. `3/2`

Answer

To solve the problem step by step, we start with the given equations and relationships.

### Step 1: Set up the equations
We know that:
1. \( \alpha + \beta = 90^\circ \)
2. \( \alpha = 2\beta \)

### Step 2: Substitute \( \alpha \) in the first equation
Substituting \( \alpha = 2\beta \) into the first equation gives us:
\[ 2\beta + \beta = 90^\circ \]
This simplifies to:
\[ 3\beta = 90^\circ \]

### Step 3: Solve for \( \beta \)
Dividing both sides by 3, we find:
\[ \beta = 30^\circ \]

### Step 4: Find \( \alpha \)
Now, substituting \( \beta \) back into the equation for \( \alpha \):
\[ \alpha = 2\beta = 2 \times 30^\circ = 60^\circ \]

### Step 5: Calculate \( 3\cos^2\alpha - 2\sin^2\beta \)
Now we need to calculate \( 3\cos^2\alpha - 2\sin^2\beta \):
- First, find \( \cos^2\alpha \) and \( \sin^2\beta \):
- \( \cos(60^\circ) = \frac{1}{2} \) so \( \cos^2(60^\circ) = \left(\frac{1}{2}\right)^2 = \frac{1}{4} \)
- \( \sin(30^\circ) = \frac{1}{2} \) so \( \sin^2(30^\circ) = \left(\frac{1}{2}\right)^2 = \frac{1}{4} \)

Now substitute these values into the expression:
\[ 3\cos^2\alpha - 2\sin^2\beta = 3\left(\frac{1}{4}\right) - 2\left(\frac{1}{4}\right) \]

### Step 6: Simplify the expression
Calculating this gives:
\[ = \frac{3}{4} - \frac{2}{4} = \frac{1}{4} \]

### Final Answer
Thus, the value of \( 3\cos^2\alpha - 2\sin^2\beta \) is:
\[ \frac{1}{4} \]