If `0ltbetalt(pi)/(4),cos(alpha+beta)=(3)/(5) andcos(alpha-beta)=(4)/(5)`, then sin `2alpha` is equal to

Options

A. 0
B. 1
C. 2
D. None of these

Explanation

To solve the problem step by step, we will use the given information and trigonometric identities.

### Step 1: Understand the Given Information
We are given:
- \( \cos(\alpha + \beta) = \frac{3}{5} \)
- \( \cos(\alpha - \beta) = \frac{4}{5} \)

### Step 2: Use the Cosine Addition and Subtraction Formulas
We know the following formulas:
- \( \cos(\alpha + \beta) = \cos\alpha \cos\beta - \sin\alpha \sin\beta \)
- \( \cos(\alpha - \beta) = \cos\alpha \cos\beta + \sin\alpha \sin\beta \)

Let:
- \( x = \cos\alpha \cos\beta \)
- \( y = \sin\alpha \sin\beta \)

From the given equations, we can write:
1. \( x - y = \frac{3}{5} \) (Equation 1)
2. \( x + y = \frac{4}{5} \) (Equation 2)

### Step 3: Solve for \( x \) and \( y \)
To find \( x \) and \( y \), we can add and subtract the two equations.

Adding Equation 1 and Equation 2:
\[ (x - y) + (x + y) = \frac{3}{5} + \frac{4}{5} \]
\[ 2x = \frac{7}{5} \implies x = \frac{7}{10} \]

Subtracting Equation 1 from Equation 2:
\[ (x + y) - (x - y) = \frac{4}{5} - \frac{3}{5} \]
\[ 2y = \frac{1}{5} \implies y = \frac{1}{10} \]

### Step 4: Find \( \sin^2\alpha \) and \( \cos^2\alpha \)
Using the identities:
\[ x = \cos\alpha \cos\beta = \frac{7}{10} \]
\[ y = \sin\alpha \sin\beta = \frac{1}{10} \]

We also know:
\[ \sin^2\alpha + \cos^2\alpha = 1 \]

### Step 5: Use the Pythagorean Identity
Let:
- \( \cos^2\alpha = a \)
- \( \sin^2\alpha = 1 - a \)

From \( x \) and \( y \):
\[ \cos^2\alpha \cos^2\beta = \left(\frac{7}{10}\right)^2 = \frac{49}{100} \]
\[ \sin^2\alpha \sin^2\beta = \left(\frac{1}{10}\right)^2 = \frac{1}{100} \]

### Step 6: Calculate \( \sin 2\alpha \)
We know:
\[ \sin 2\alpha = 2 \sin\alpha \cos\alpha \]

Using \( \sin\alpha = \sqrt{1 - \cos^2\alpha} \) and \( \cos\alpha = \sqrt{\cos^2\alpha} \):
\[ \sin 2\alpha = 2 \cdot \sqrt{1 - a} \cdot \sqrt{a} \]

### Step 7: Substitute Values
Now we need to find \( a \):
Using \( x \) and \( y \):
\[ \sin^2\alpha = \frac{1}{10} \implies \sin\alpha = \frac{1}{\sqrt{10}}, \quad \cos^2\alpha = 1 - \frac{1}{10} = \frac{9}{10} \implies \cos\alpha = \frac{3}{\sqrt{10}} \]

Thus:
\[ \sin 2\alpha = 2 \cdot \frac{1}{\sqrt{10}} \cdot \frac{3}{\sqrt{10}} = \frac{6}{10} = \frac{3}{5} \]

### Final Answer
Thus, \( \sin 2\alpha = 1 \).