If `(sin(A+B))/(cos(A-B))=(1-m)/(1+m)`,then `tan(pi/4-A)tan(pi/4-B)=?`

Options

A. m
B. 2m
C. 3m
D. 4m

Answer

To solve the problem, we start with the given equation:

\[ \frac{\sin(A+B)}{\cos(A-B)} = \frac{1-m}{1+m} \]

We need to find the value of \( \tan\left(\frac{\pi}{4} - A\right) \tan\left(\frac{\pi}{4} - B\right) \).

### Step 1: Rewrite the left-hand side using trigonometric identities

Using the identities for sine and cosine, we can rewrite the left-hand side:

\[ \sin(A+B) = \sin A \cos B + \cos A \sin B \]
\[ \cos(A-B) = \cos A \cos B + \sin A \sin B \]

Thus, we have:

\[ \frac{\sin A \cos B + \cos A \sin B}{\cos A \cos B + \sin A \sin B} = \frac{1-m}{1+m} \]

### Step 2: Cross-multiply to eliminate the fraction

Cross-multiplying gives us:

\[ (1-m)(\cos A \cos B + \sin A \sin B) = (1+m)(\sin A \cos B + \cos A \sin B) \]

### Step 3: Expand both sides

Expanding both sides, we get:

\[ \cos A \cos B + \sin A \sin B - m(\cos A \cos B + \sin A \sin B) = \sin A \cos B + \cos A \sin B + m(\sin A \cos B + \cos A \sin B) \]

### Step 4: Rearranging the equation

Rearranging gives us:

\[ \cos A \cos B + \sin A \sin B - \sin A \cos B - \cos A \sin B = m(\sin A \cos B + \cos A \sin B + \cos A \cos B + \sin A \sin B) \]

### Step 5: Factor the left-hand side

The left-hand side can be factored as:

\[ (\cos A - \sin B)(\cos B - \sin A) = m(\sin A + \cos A)(\sin B + \cos B) \]

### Step 6: Find \( \tan\left(\frac{\pi}{4} - A\right) \tan\left(\frac{\pi}{4} - B\right) \)

Using the identity:

\[ \tan\left(\frac{\pi}{4} - A\right) = \frac{1 - \tan A}{1 + \tan A} \]
\[ \tan\left(\frac{\pi}{4} - B\right) = \frac{1 - \tan B}{1 + \tan B} \]

Thus, we have:

\[ \tan\left(\frac{\pi}{4} - A\right) \tan\left(\frac{\pi}{4} - B\right) = \frac{(1 - \tan A)(1 - \tan B)}{(1 + \tan A)(1 + \tan B)} \]

### Step 7: Substitute values and simplify

Let \( x = \tan A \) and \( y = \tan B \). Then we can rewrite:

\[ \tan\left(\frac{\pi}{4} - A\right) \tan\left(\frac{\pi}{4} - B\right) = \frac{(1 - x)(1 - y)}{(1 + x)(1 + y)} \]

### Final Result

After simplification, we find that:

\[ \tan\left(\frac{\pi}{4} - A\right) \tan\left(\frac{\pi}{4} - B\right) = \frac{(1 - m)(1 - m)}{(1 + m)(1 + m)} = \frac{(1 - m)^2}{(1 + m)^2} \]

Thus, the final answer is:

\[ \tan\left(\frac{\pi}{4} - A\right) \tan\left(\frac{\pi}{4} - B\right) = \frac{(1 - m)^2}{(1 + m)^2} \]