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

Ans: We To find the value of tan A + tan B, we need to find the value of tan A and tan B

Step 1: We are given sin A = \frac{3}{5}

We know that sin2 A + cos2 A = 1

(\frac{3}{5})^2 + cos2 A = 1

∴ cos2 A = 1 – \frac{9}{25} = \frac{16}{25}

∴ cos A = \sqrt{\frac{16}{25}} = \frac{4}{5}

Now, tan A = \frac{\sin A}{\cos A}

By substituting value of sin A and cos A in above equation, we get:

tan A = \frac{\frac{3}{5}}{\frac{4}{5}} = \frac{3}{4}

Step 2: Next, we have cos B = \frac{12}{13}

∵ sin2 B + cos2 B = 1

∴ sin2 B +  (\frac{12}{13})^2 = 1

∴ sin2 B = 1 – \frac{144}{169} = \frac{25}{169}

∴ sin B = \sqrt{\frac{25}{169}} = \frac{5}{13}

Now, tan B = \frac{\sin B}{\cos B}

By substituting value of sin B and cos B in above equation, we get:

tan A = \frac{\frac{5}{13}}{\frac{12}{13}} = \frac{5}{12}

Step 3: Let’s find the value of (tan A + tan B) by substituting the values from step 1 and step 2:

(tan A + tan B) = \frac{3}{4} + \frac{5}{12} = \frac{9 + 5}{12}

=\frac{14}{12} = \frac{7}{6} = 1 \frac{1}{6}

Therefore, the value of (tan A + tan B) is 1 \frac{1}{6}

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