Question

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

Answer 1

We know:
$\tan A = \frac{\sin A}{\cos A}, \quad \tan B = \frac{\sin B}{\cos B}.$
Step 1: Find $\cos A$ Using the Pythagorean identity:
$\sin^2 A + \cos^2 A = 1 \implies \cos^2 A = 1 - \sin^2 A = 1 - \left(\frac{3}{5}\right)^2 = \frac{16}{25}.$
$\cos A = \frac{4}{5}.$
Step 2: Find $\sin B$ Using the Pythagorean identity:
$\sin^2 B + \cos^2 B = 1 \implies \sin^2 B = 1 - \cos^2 B = 1 - \left(\frac{12}{13}\right)^2 = \frac{25}{169}.$
$\sin B = \frac{5}{13}.$
Step 3: Calculate $\tan A$ and $\tan B$
$\tan A = \frac{\sin A}{\cos A} = \frac{\frac{3}{5}}{\frac{4}{5}} = \frac{3}{4}, \quad \tan B = \frac{\sin B}{\cos B} = \frac{\frac{5}{13}}{\frac{12}{13}} = \frac{5}{12}.$
Step 4: Find $(\tan A + \tan B)$
$\tan A + \tan B = \frac{3}{4} + \frac{5}{12}.$
Taking the LCM of 4 and 12:
$\tan A + \tan B = \frac{9}{12} + \frac{5}{12} = \frac{14}{12} = \frac{7}{6}.$
Correct Answer: $\frac{7}{6}$.

Answer 2

We know:
$\tan A = \frac{\sin A}{\cos A}, \quad \tan B = \frac{\sin B}{\cos B}.$
Step 1: Find $\cos A$ Using the Pythagorean identity:
$\sin^2 A + \cos^2 A = 1 \implies \cos^2 A = 1 - \sin^2 A = 1 - \left(\frac{3}{5}\right)^2 = \frac{16}{25}.$
$\cos A = \frac{4}{5}.$
Step 2: Find $\sin B$ Using the Pythagorean identity:
$\sin^2 B + \cos^2 B = 1 \implies \sin^2 B = 1 - \cos^2 B = 1 - \left(\frac{12}{13}\right)^2 = \frac{25}{169}.$
$\sin B = \frac{5}{13}.$
Step 3: Calculate $\tan A$ and $\tan B$
$\tan A = \frac{\sin A}{\cos A} = \frac{\frac{3}{5}}{\frac{4}{5}} = \frac{3}{4}, \quad \tan B = \frac{\sin B}{\cos B} = \frac{\frac{5}{13}}{\frac{12}{13}} = \frac{5}{12}.$
Step 4: Find $(\tan A + \tan B)$
$\tan A + \tan B = \frac{3}{4} + \frac{5}{12}.$
Taking the LCM of 4 and 12:
$\tan A + \tan B = \frac{9}{12} + \frac{5}{12} = \frac{14}{12} = \frac{7}{6}.$
Correct Answer: $\frac{7}{6}$.