If `sinA=(3)/(5)andcosB=(12)/(13)`,what is the value of `(tanA-tanB)/(1+tanAtanB)`,it being given that A and B are acute angles ?

Options

A. `(15)/(63)`
B. `(19)/(63)`
C. `(17)/(65)`
D. `(16)/(63)`

Answer

To solve the problem, we need to find the value of \((\tan A - \tan B) / (1 + \tan A \tan B)\) given that \(\sin A = \frac{3}{5}\) and \(\cos B = \frac{12}{13}\).

### Step 1: Find \(\tan A\)

Given \(\sin A = \frac{3}{5}\), we can use the Pythagorean identity to find \(\cos A\):
\[ \cos^2 A + \sin^2 A = 1 \]
Substituting \(\sin A\):
\[ \cos^2 A + \left(\frac{3}{5}\right)^2 = 1 \]
\[ \cos^2 A + \frac{9}{25} = 1 \]
\[ \cos^2 A = 1 - \frac{9}{25} = \frac{16}{25} \]
\[ \cos A = \sqrt{\frac{16}{25}} = \frac{4}{5} \]
Now, we can find \(\tan A\):
\[ \tan A = \frac{\sin A}{\cos A} = \frac{\frac{3}{5}}{\frac{4}{5}} = \frac{3}{4} \]

### Step 2: Find \(\tan B\)

Given \(\cos B = \frac{12}{13}\), we can use the Pythagorean identity to find \(\sin B\):
\[ \sin^2 B + \cos^2 B = 1 \]
Substituting \(\cos B\):
\[ \sin^2 B + \left(\frac{12}{13}\right)^2 = 1 \]
\[ \sin^2 B + \frac{144}{169} = 1 \]
\[ \sin^2 B = 1 - \frac{144}{169} = \frac{25}{169} \]
\[ \sin B = \sqrt{\frac{25}{169}} = \frac{5}{13} \]
Now, we can find \(\tan B\):
\[ \tan B = \frac{\sin B}{\cos B} = \frac{\frac{5}{13}}{\frac{12}{13}} = \frac{5}{12} \]

### Step 3: Substitute \(\tan A\) and \(\tan B\) into the expression

Now we substitute \(\tan A\) and \(\tan B\) into the expression:
\[ \frac{\tan A - \tan B}{1 + \tan A \tan B} = \frac{\frac{3}{4} - \frac{5}{12}}{1 + \frac{3}{4} \cdot \frac{5}{12}} \]

### Step 4: Simplify the numerator

To simplify \(\frac{3}{4} - \frac{5}{12}\), we need a common denominator:
The least common multiple of 4 and 12 is 12.
\[ \frac{3}{4} = \frac{9}{12} \]
So,
\[ \frac{3}{4} - \frac{5}{12} = \frac{9}{12} - \frac{5}{12} = \frac{4}{12} = \frac{1}{3} \]

### Step 5: Simplify the denominator

Now we simplify \(1 + \tan A \tan B\):
\[ \tan A \tan B = \frac{3}{4} \cdot \frac{5}{12} = \frac{15}{48} \]
Now, simplifying \(1 + \frac{15}{48}\):
\[ 1 = \frac{48}{48} \]
Thus,
\[ 1 + \frac{15}{48} = \frac{48 + 15}{48} = \frac{63}{48} \]

### Step 6: Combine the results

Now we can combine the results:
\[ \frac{\frac{1}{3}}{\frac{63}{48}} = \frac{1}{3} \cdot \frac{48}{63} = \frac{48}{189} \]

### Step 7: Simplify the fraction

We can simplify \(\frac{48}{189}\):
\[ \frac{48 \div 3}{189 \div 3} = \frac{16}{63} \]

Thus, the final answer is:
\[ \frac{\tan A - \tan B}{1 + \tan A \tan B} = \frac{16}{63} \]