(d) \( \sin \mathrm{A}=\frac{12}{13} \) and \( \mathrm{A} \in\left(90^{\circ} ; 270^{\circ}\right) \), determine without the use of a calculator and with the aid of a diagram the value of: \( \begin{array}{ll}\text { (1) } \tan \mathrm{A} & \text { (2) } \text { (d }-\sin ^{2} \mathrm{~A}\end{array} \)

To start, since \( \sin \mathrm{A} = \frac{12}{13} \) and angle \( A \) lies in the range \( (90^{\circ}; 270^{\circ) \), we know it is in the second or third quadrant where sine is positive only in the second quadrant. In this case, we can determine \( \cos A \) using the Pythagorean identity \( \sin^2 A + \cos^2 A = 1 \). 1. First, calculate \( \cos A \): \[ \sin^2 \mathrm{A} = \left(\frac{12}{13}\right)^2 = \frac{144}{169} \] So, \[ \cos^2 \mathrm{A} = 1 - \sin^2 \mathrm{A} = 1 - \frac{144}{169} = \frac{169 - 144}{169} = \frac{25}{169} \] Therefore, \( \cos A = -\frac{5}{13} \) (negative in the second quadrant). Now, we can find \( \tan A \): \[ \tan \mathrm{A} = \frac{\sin \mathrm{A}}{\cos \mathrm{A}} = \frac{\frac{12}{13}}{-\frac{5}{13}} = -\frac{12}{5} \] 2. Next, we need to determine \( d - \sin^2 A \): Since \( d = 1 \) (the identity for sine), we have: \[ d - \sin^2 \mathrm{A} = 1 - \frac{144}{169} = \frac{169}{169} - \frac{144}{169} = \frac{25}{169} \] So, the results are: \[ \text{(1) } \tan \mathrm{A} = -\frac{12}{5} \] \[ \text{(2) } d - \sin^2 \mathrm{A} = \frac{25}{169} \]