(d) If \( \sin \mathrm{A}=\frac{12}{13} \) 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: \( \begin{array}{lll}\text { (1) } \tan \mathrm{A} & \text { (2) } \cos ^{2} \mathrm{~A}-\sin ^{2} \mathrm{~A} & \text { (3) } 13 \cos \mathrm{~A}+5 \tan \mathrm{~A} \\ \text { (e) If } \cos \mathrm{B}=\frac{-8}{17} \text { and } \mathrm{B} \in\left(180^{\circ} ; 360^{\circ}\right) \text {, determine without the use of a calculator and with the } \\ \text { aid of a diagram the value of: } \\ \begin{array}{lll}\text { (1) } \sin \mathrm{B} & \text { (2) } \frac{1}{\tan ^{2} \mathrm{~B}}-\frac{1}{\sin ^{2} \mathrm{~B}} & \text { (3) } 17 \sin \mathrm{~B}-8 \tan \mathrm{~B}\end{array}\end{array}> \)

**(d) Given**: \[ \sin A=\frac{12}{13},\quad A\in\left(90^{\circ},270^{\circ}\right) \] Since \(\sin A > 0\) and \(A\) lies between \(90^{\circ}\) and \(270^{\circ}\), it must be in the second quadrant. In the second quadrant, \(\cos A\) is negative. 1. **Find \(\cos A\)**: Using the identity \[ \sin^2 A+\cos^2 A=1, \] we have \[ \cos^2 A=1-\sin^2 A=1-\left(\frac{12}{13}\right)^2=1-\frac{144}{169}=\frac{25}{169}. \] Since \(A\) is in the second quadrant, \[ \cos A=-\frac{5}{13}. \] 2. **(1) Compute \(\tan A\)**: \[ \tan A=\frac{\sin A}{\cos A}=\frac{\frac{12}{13}}{-\frac{5}{13}}=-\frac{12}{5}. \] 3. **(2) Compute \(\cos^2A-\sin^2A\)**: \[ \cos^2A-\sin^2A=\frac{25}{169}-\frac{144}{169}=-\frac{119}{169}. \] 4. **(3) Compute \(13\cos A+5\tan A\)**: \[ 13\cos A+5\tan A=13\left(-\frac{5}{13}\right)+5\left(-\frac{12}{5}\right)=-5-12=-17. \] --- **(e) Given**: \[ \cos B=\frac{-8}{17},\quad B\in\left(180^{\circ},360^{\circ}\right) \] For angles in this interval, cosine is negative in the third quadrant and positive only in the fourth. Since \(\cos B\) is negative, \(B\) is in the third quadrant where both sine and cosine are negative. 1. **Find \(\sin B\)**: Using the Pythagorean identity, \[ \sin^2B=1-\cos^2B=1-\left(\frac{-8}{17}\right)^2=1-\frac{64}{289}=\frac{225}{289}. \] In the third quadrant, \[ \sin B=-\frac{15}{17}. \] 2. **(1) \(\sin B\)** is already found: \[ \sin B=-\frac{15}{17}. \] 3. **Find \(\tan B\)**: \[ \tan B=\frac{\sin B}{\cos B}=\frac{-\frac{15}{17}}{-\frac{8}{17}}=\frac{15}{8}. \] 4. **(2) Compute \(\frac{1}{\tan^2B}-\frac{1}{\sin^2B}\)**: First, calculate \(\tan^2B\) and \(\sin^2B\): \[ \tan^2B=\left(\frac{15}{8}\right)^2=\frac{225}{64},\quad \sin^2B=\frac{225}{289}. \] Then, \[ \frac{1}{\tan^2B}=\frac{64}{225},\quad \frac{1}{\sin^2B}=\frac{289}{225}. \] So, \[ \frac{1}{\tan^2B}-\frac{1}{\sin^2B}=\frac{64}{225}-\frac{289}{225}=-\frac{225}{225}=-1. \] 5. **(3) Compute \(17\sin B-8\tan B\)**: \[ 17\sin B-8\tan B=17\left(-\frac{15}{17}\right)-8\left(\frac{15}{8}\right)=-15-15=-30. \] --- **Final Answers**: **(d)** 1. \(\tan A=-\frac{12}{5}\) 2. \(\cos^2 A-\sin^2 A=-\frac{119}{169}\) 3. \(13\cos A+5\tan A=-17\) **(e)** 1. \(\sin B=-\frac{15}{17}\) 2. \(\frac{1}{\tan^2B}-\frac{1}{\sin^2B}=-1\) 3. \(17\sin B-8\tan B=-30\)