(f) If $$ 4 an B-3=0 $$ and $$ \cos B

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

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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:** $$ \text{(1)} \quad (\sin B+\cos B)^2 = \frac{49}{25}, $$ $$ \text{(2)} \quad 25(\sin B-\cos B)^2 = 1. $$ $$ \text{(1)} \quad (\sin B + \cos B)^2 = \frac{49}{25}, $$ $$ \text{(2)} \quad 25(\sin B - \cos B)^2 = 1. $$

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

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