If `cosA=(3)/(5) and sinB=(5)/(13)`, find the value of `(tanA+tanB)/(1-tanAtanB)` :

Options

A. `(63)/(16)`
B. `(36)/(16)`
C. `(61)/(36)`
D. none of these

Answer

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

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

Given \(\cos A = \frac{3}{5}\), we can find \(\sin A\) using the Pythagorean identity:
\[ \sin^2 A + \cos^2 A = 1 \]
Substituting the value of \(\cos A\):
\[ \sin^2 A + \left(\frac{3}{5}\right)^2 = 1 \]
\[ \sin^2 A + \frac{9}{25} = 1 \]
\[ \sin^2 A = 1 - \frac{9}{25} = \frac{25}{25} - \frac{9}{25} = \frac{16}{25} \]
Taking the square root:
\[ \sin A = \frac{4}{5} \quad (\text{since } A \text{ is in the first quadrant}) \]
Now, we can find \(\tan A\):
\[ \tan A = \frac{\sin A}{\cos A} = \frac{\frac{4}{5}}{\frac{3}{5}} = \frac{4}{3} \]

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

Given \(\sin B = \frac{5}{13}\), we can find \(\cos B\) using the Pythagorean identity:
\[ \sin^2 B + \cos^2 B = 1 \]
Substituting the value of \(\sin B\):
\[ \left(\frac{5}{13}\right)^2 + \cos^2 B = 1 \]
\[ \frac{25}{169} + \cos^2 B = 1 \]
\[ \cos^2 B = 1 - \frac{25}{169} = \frac{169}{169} - \frac{25}{169} = \frac{144}{169} \]
Taking the square root:
\[ \cos B = \frac{12}{13} \quad (\text{since } B \text{ is in the first quadrant}) \]
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:
\[ \frac{\tan A + \tan B}{1 - \tan A \tan B} = \frac{\frac{4}{3} + \frac{5}{12}}{1 - \left(\frac{4}{3} \cdot \frac{5}{12}\right)} \]

### Step 4: Simplify the numerator

To add \(\frac{4}{3}\) and \(\frac{5}{12}\), we need a common denominator:
\[ \frac{4}{3} = \frac{16}{12} \]
Thus,
\[ \frac{4}{3} + \frac{5}{12} = \frac{16}{12} + \frac{5}{12} = \frac{21}{12} \]

### Step 5: Simplify the denominator

Now calculate \(\tan A \tan B\):
\[ \tan A \tan B = \frac{4}{3} \cdot \frac{5}{12} = \frac{20}{36} = \frac{5}{9} \]
Now substitute into the denominator:
\[ 1 - \tan A \tan B = 1 - \frac{5}{9} = \frac{9}{9} - \frac{5}{9} = \frac{4}{9} \]

### Step 6: Final calculation

Now substituting back into the expression:
\[ \frac{\frac{21}{12}}{\frac{4}{9}} = \frac{21}{12} \cdot \frac{9}{4} = \frac{21 \cdot 9}{12 \cdot 4} = \frac{189}{48} \]
Now simplify \(\frac{189}{48}\):
\[ \frac{189 \div 3}{48 \div 3} = \frac{63}{16} \]

Thus, the final answer is:
\[ \frac{63}{16} \]