If `A+B=45^(@),"then "(tanA)/(1-tanA).(tanB)/(1-tanB)=` ?

Answer

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

**Given:** \( A + B = 45^\circ \)

We need to find the value of:

\[ \frac{\tan A}{1 - \tan A} \cdot \frac{\tan B}{1 - \tan B} \]

### Step 1: Use the identity for \( A + B = 45^\circ \)

Since \( A + B = 45^\circ \), we can use the tangent addition formula:

\[ \tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} \]

Substituting \( A + B = 45^\circ \):

\[ \tan 45^\circ = 1 \]

So we have:

\[ 1 = \frac{\tan A + \tan B}{1 - \tan A \tan B} \]

### Step 2: Rearranging the equation

From the equation above, we can rearrange it to find:

\[ 1 - \tan A \tan B = \tan A + \tan B \]

### Step 3: Express the product

Now, we can express the product we need to find:

\[ \frac{\tan A}{1 - \tan A} \cdot \frac{\tan B}{1 - \tan B} \]

### Step 4: Substitute \( 1 - \tan A \) and \( 1 - \tan B \)

We can rewrite \( \frac{\tan A}{1 - \tan A} \) and \( \frac{\tan B}{1 - \tan B} \):

\[ \frac{\tan A}{1 - \tan A} = \frac{\tan A}{\tan A + \tan B} \]

\[ \frac{\tan B}{1 - \tan B} = \frac{\tan B}{\tan A + \tan B} \]

### Step 5: Combine the fractions

Now we can combine the fractions:

\[ \frac{\tan A \tan B}{(1 - \tan A)(1 - \tan B)} = \frac{\tan A \tan B}{\tan A + \tan B} \]

### Step 6: Use the identity

From the previous steps, we know that:

\[ 1 - \tan A \tan B = \tan A + \tan B \]

Thus, we can conclude:

\[ \frac{\tan A \tan B}{\tan A + \tan B} = \frac{1}{2} \]

### Final Result

So, the value of:

\[ \frac{\tan A}{1 - \tan A} \cdot \frac{\tan B}{1 - \tan B} = \frac{1}{2} \]