The angles \( A \) and \( B \) are such that \( \sin A=\frac{3}{5} \) and \( \cos B=-\frac{4}{5} \), where \( A \) and \( B \) are in the same quadrant. Find the value of each of the following without using a calculator. \( \begin{array}{ll}\text { (i) } \tan (A+B) & \text { (ii) } \cos (A+B) \\ \text { (iii) } \sin (A-B) & \end{array} \)

Given that both angles \( A \) and \( B \) are in the same quadrant and: \[ \sin A = \frac{3}{5}, \quad \cos B = -\frac{4}{5} \] ### Determining the Quadrant: - \( \sin A = \frac{3}{5} > 0 \) implies that angle \( A \) is either in the **First** or **Second** quadrant. - \( \cos B = -\frac{4}{5} < 0 \) implies that angle \( B \) is either in the **Second** or **Third** quadrant. Since \( A \) and \( B \) are in the **same quadrant**, both must be in the **Second** quadrant. ### Finding Remaining Trigonometric Values: For angle \( A \) in the Second Quadrant: \[ \cos A = -\sqrt{1 - \sin^2 A} = -\sqrt{1 - \left(\frac{3}{5}\right)^2} = -\frac{4}{5} \] For angle \( B \) in the Second Quadrant: \[ \sin B = \sqrt{1 - \cos^2 B} = \sqrt{1 - \left(-\frac{4}{5}\right)^2} = \frac{3}{5} \] ### (i) \( \tan(A + B) \): Using the tangent addition formula: \[ \tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} \] First, calculate \( \tan A \) and \( \tan B \): \[ \tan A = \frac{\sin A}{\cos A} = \frac{\frac{3}{5}}{-\frac{4}{5}} = -\frac{3}{4} \] \[ \tan B = \frac{\sin B}{\cos B} = \frac{\frac{3}{5}}{-\frac{4}{5}} = -\frac{3}{4} \] Now, substitute into the addition formula: \[ \tan(A + B) = \frac{-\frac{3}{4} - \frac{3}{4}}{1 - \left(-\frac{3}{4}\right)\left(-\frac{3}{4}\right)} = \frac{-\frac{6}{4}}{1 - \frac{9}{16}} = \frac{-\frac{3}{2}}{\frac{7}{16}} = -\frac{24}{7} \] ### (ii) \( \cos(A + B) \): Using the cosine addition formula: \[ \cos(A + B) = \cos A \cos B - \sin A \sin B \] \[ \cos(A + B) = \left(-\frac{4}{5}\right)\left(-\frac{4}{5}\right) - \left(\frac{3}{5}\right)\left(\frac{3}{5}\right) = \frac{16}{25} - \frac{9}{25} = \frac{7}{25} \] ### (iii) \( \sin(A - B) \): Using the sine subtraction formula: \[ \sin(A - B) = \sin A \cos B - \cos A \sin B \] \[ \sin(A - B) = \left(\frac{3}{5}\right)\left(-\frac{4}{5}\right) - \left(-\frac{4}{5}\right)\left(\frac{3}{5}\right) = -\frac{12}{25} + \frac{12}{25} = 0 \] ### Summary of Results: \[ \begin{aligned} &\text{(i)} \quad \tan(A + B) = -\frac{24}{7} \\ &\text{(ii)} \quad \cos(A + B) = \frac{7}{25} \\ &\text{(iii)} \quad \sin(A - B) = 0 \\ \end{aligned} \] **Answer:** (i) \( \tan(A+B) = -\dfrac{24}{7} \) (ii) \( \cos(A+B) = \dfrac{7}{25} \) (iii) \( \sin(A-B) = 0 \)