Question

If $\sin A = \frac{3}{5}$ and $\cos B = \frac{12}{13}$, then find the value of $(\tan A + \tan B)$.

Answer 1

Step 1: Use the given information:
We are given: - \( \sin A = \frac{3}{5} \) - \( \cos B = \frac{12}{13} \) We are asked to find the value of \( \tan A + \tan B \).

Step 2: Find \( \tan A \):
We know that \( \tan A = \frac{\sin A}{\cos A} \). To find \( \tan A \), we first need to find \( \cos A \). We can use the Pythagorean identity \( \sin^2 A + \cos^2 A = 1 \). Substituting \( \sin A = \frac{3}{5} \) into this identity:
\[ \left( \frac{3}{5} \right)^2 + \cos^2 A = 1 \] \[ \frac{9}{25} + \cos^2 A = 1 \] \[ \cos^2 A = 1 - \frac{9}{25} = \frac{25}{25} - \frac{9}{25} = \frac{16}{25} \] \[ \cos A = \frac{4}{5} \] Now we can find \( \tan A \):
\[ \tan A = \frac{\sin A}{\cos A} = \frac{\frac{3}{5}}{\frac{4}{5}} = \frac{3}{4} \]

Step 3: Find \( \tan B \):
We know that \( \tan B = \frac{\sin B}{\cos B} \). To find \( \tan B \), we first need to find \( \sin B \). Again, using the Pythagorean identity \( \sin^2 B + \cos^2 B = 1 \), we substitute \( \cos B = \frac{12}{13} \):
\[ \left( \frac{12}{13} \right)^2 + \sin^2 B = 1 \] \[ \frac{144}{169} + \sin^2 B = 1 \] \[ \sin^2 B = 1 - \frac{144}{169} = \frac{169}{169} - \frac{144}{169} = \frac{25}{169} \] \[ \sin B = \frac{5}{13} \] Now we can find \( \tan B \):
\[ \tan B = \frac{\sin B}{\cos B} = \frac{\frac{5}{13}}{\frac{12}{13}} = \frac{5}{12} \]

Step 4: Find \( \tan A + \tan B \):
Now that we have \( \tan A = \frac{3}{4} \) and \( \tan B = \frac{5}{12} \), we can find \( \tan A + \tan B \):
\[ \tan A + \tan B = \frac{3}{4} + \frac{5}{12} \] To add these fractions, we need a common denominator. The least common denominator of 4 and 12 is 12, so we rewrite \( \frac{3}{4} \) as \( \frac{9}{12} \):
\[ \tan A + \tan B = \frac{9}{12} + \frac{5}{12} = \frac{14}{12} = \frac{7}{6} \]

Conclusion:
The value of \( \tan A + \tan B \) is \( \frac{7}{6} \).