(f) If \( 4 \tan B-3=0 \) and \( \cos B<0 \), determine by means of a diagram: \( \begin{array}{ll}\text { (1) }(\sin B+\cos B)^{2} & \text { (2) } 25(\sin B-\cos B)^{2}\end{array} \)

We start with the equation \[ 4\tan B - 3 = 0. \] **Step 1. Solve for \(\tan B\).** \[ 4\tan B = 3 \quad \Longrightarrow \quad \tan B = \frac{3}{4}. \] **Step 2. Decide the quadrant for angle \(B\).** Since we are given that \[ \cos B < 0, \] and knowing that \[ \tan B = \frac{\sin B}{\cos B}, \] a positive \(\tan B\) (because \(\frac{3}{4} > 0\)) implies that \(\sin B\) and \(\cos B\) must have the same sign. With \(\cos B < 0\), it follows that \(\sin B\) must also be negative if both were in the third quadrant. However, in the third quadrant \(\tan\) is positive, but then both sine and cosine are negative. Thus, there are two possibilities: - \(\tan B\) positive in the first quadrant, but then \(\cos B\) would be positive. - \(\tan B\) positive in the third quadrant, where both \(\sin B\) and \(\cos B\) are negative. Yet, the diagram clue often indicates drawing a reference triangle where the legs are taken in magnitude. We consider the reference triangle with legs 3 and 4. Then, since \(\tan B = \frac{3}{4}\), we set: - Opposite side = 3, - Adjacent side = 4, - Hypotenuse = \(\sqrt{3^2+4^2} = 5\). Now, because \(\cos B<0\) and the reference triangle gives the positive values, the actual cosine must be negative. At this point we have two common situations: 1. Third Quadrant: \(\sin B < 0\) and \(\cos B < 0\) \(\Longrightarrow\) \(\tan B\) positive. 2. Second Quadrant: \(\sin B > 0\) and \(\cos B < 0\) \(\Longrightarrow\) \(\tan B\) negative. The correct quadrant with \(\cos B < 0\) and \(\tan B =\frac{3}{4}>0\) is therefore the third quadrant (where both sine and cosine are negative). Thus we have \[ \sin B = -\frac{3}{5} \quad \text{and} \quad \cos B = -\frac{4}{5}. \] **Step 3. Calculate \((\sin B+\cos B)^2\).** \[ \sin B + \cos B = -\frac{3}{5} - \frac{4}{5} = -\frac{7}{5}. \] Then \[ (\sin B+\cos B)^2 = \left(-\frac{7}{5}\right)^2 = \frac{49}{25}. \] **Step 4. Calculate \(25(\sin B-\cos B)^2\).** \[ \sin B - \cos B = -\frac{3}{5} -\left(-\frac{4}{5}\right) = -\frac{3}{5} + \frac{4}{5} = \frac{1}{5}. \] So \[ (\sin B-\cos B)^2 = \left(\frac{1}{5}\right)^2 = \frac{1}{25}, \] and therefore \[ 25(\sin B-\cos B)^2 = 25 \times \frac{1}{25} = 1. \] **Diagram Explanation:** - Draw a right triangle where one leg (representing the opposite side) is 3 and the other leg (representing the adjacent side) is 4. - The hypotenuse is 5. - Since angle \(B\) is located in the third quadrant, both \(\sin B\) and \(\cos B\) are taken as negative: - \(\sin B = -\frac{3}{5}\) - \(\cos B = -\frac{4}{5}\) **Final Answers:** \[ \text{(1)} \quad (\sin B+\cos B)^2 = \frac{49}{25}, \] \[ \text{(2)} \quad 25(\sin B-\cos B)^2 = 1. \]