If A is an obtuse angle whose sine is `(5)/(13) and B ` is an acute angle whose tangent is `(3)/(4),` without using tables find the values of
(a) `sin 2B,`
(b)` tan (A-B).`

Answer

To solve the problem, we need to find the values of \( \sin 2B \) and \( \tan(A - B) \) given that \( \sin A = \frac{5}{13} \) (where \( A \) is an obtuse angle) and \( \tan B = \frac{3}{4} \) (where \( B \) is an acute angle).

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

Since \( A \) is an obtuse angle, we can find \( \cos A \) using the identity:

\[ \cos^2 A + \sin^2 A = 1 \]

Substituting \( \sin A = \frac{5}{13} \):

\[ \cos^2 A + \left(\frac{5}{13}\right)^2 = 1 \]

Calculating \( \left(\frac{5}{13}\right)^2 \):

\[ \cos^2 A + \frac{25}{169} = 1 \]

Now, subtract \( \frac{25}{169} \) from 1:

\[ \cos^2 A = 1 - \frac{25}{169} = \frac{169 - 25}{169} = \frac{144}{169} \]

Taking the square root:

\[ \cos A = -\sqrt{\frac{144}{169}} = -\frac{12}{13} \]

(The negative sign is because \( A \) is obtuse.)

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

Using the double angle formula for sine:

\[ \sin 2B = \frac{2 \tan B}{1 + \tan^2 B} \]

Substituting \( \tan B = \frac{3}{4} \):

\[ \sin 2B = \frac{2 \cdot \frac{3}{4}}{1 + \left(\frac{3}{4}\right)^2} \]

Calculating \( \left(\frac{3}{4}\right)^2 \):

\[ \sin 2B = \frac{\frac{6}{4}}{1 + \frac{9}{16}} = \frac{\frac{6}{4}}{\frac{16 + 9}{16}} = \frac{\frac{6}{4}}{\frac{25}{16}} \]

Now, multiply by the reciprocal:

\[ \sin 2B = \frac{6}{4} \cdot \frac{16}{25} = \frac{96}{100} = \frac{24}{25} \]

### Step 3: Find \( \tan A \)

Using the definition of tangent:

\[ \tan A = \frac{\sin A}{\cos A} = \frac{\frac{5}{13}}{-\frac{12}{13}} = -\frac{5}{12} \]

### Step 4: Find \( \tan(A - B) \)

Using the formula for tangent of the difference of angles:

\[ \tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B} \]

Substituting \( \tan A = -\frac{5}{12} \) and \( \tan B = \frac{3}{4} \):

\[ \tan(A - B) = \frac{-\frac{5}{12} - \frac{3}{4}}{1 + \left(-\frac{5}{12}\right) \left(\frac{3}{4}\right)} \]

Finding a common denominator for the numerator:

\[ -\frac{5}{12} - \frac{3}{4} = -\frac{5}{12} - \frac{9}{12} = -\frac{14}{12} = -\frac{7}{6} \]

Now calculating the denominator:

\[ 1 + \left(-\frac{5}{12}\right) \left(\frac{3}{4}\right) = 1 - \frac{15}{48} = \frac{48 - 15}{48} = \frac{33}{48} \]

Now substituting back into the tangent formula:

\[ \tan(A - B) = \frac{-\frac{7}{6}}{\frac{33}{48}} = -\frac{7}{6} \cdot \frac{48}{33} = -\frac{56}{33} \]

### Final Answers

(a) \( \sin 2B = \frac{24}{25} \)

(b) \( \tan(A - B) = -\frac{56}{33} \)