Answer :
To find [tex]\(\tan(A - B)\)[/tex], we start by determining the values of [tex]\(\tan(A)\)[/tex] and [tex]\(\tan(B)\)[/tex] given the provided conditions.
### Step 1: Finding [tex]\(\tan(A)\)[/tex]
Given:
[tex]\[ \sin(A) = \frac{5}{13} \][/tex]
Since [tex]\( \frac{\pi}{2} < A < \pi \)[/tex], angle [tex]\(A\)[/tex] is in the second quadrant. In the second quadrant, cosine is negative. We use the Pythagorean identity:
[tex]\[ \cos^2(A) + \sin^2(A) = 1 \][/tex]
[tex]\[ \cos^2(A) = 1 - \sin^2(A) \][/tex]
[tex]\[ \cos^2(A) = 1 - \left(\frac{5}{13}\right)^2 \][/tex]
[tex]\[ \cos^2(A) = 1 - \frac{25}{169} \][/tex]
[tex]\[ \cos^2(A) = \frac{144}{169} \][/tex]
[tex]\[ \cos(A) = -\sqrt{\frac{144}{169}} \][/tex]
[tex]\[ \cos(A) = -\frac{12}{13} \][/tex]
Now, we can find [tex]\(\tan(A)\)[/tex]:
[tex]\[ \tan(A) = \frac{\sin(A)}{\cos(A)} \][/tex]
[tex]\[ \tan(A) = \frac{\frac{5}{13}}{-\frac{12}{13}} \][/tex]
[tex]\[ \tan(A) = -\frac{5}{12} \][/tex]
### Step 2: Finding [tex]\(\tan(B)\)[/tex]
Given:
[tex]\[ \tan(B) = -\sqrt{13} \][/tex]
Since [tex]\( \frac{\pi}{2} < B < \pi \)[/tex], angle [tex]\(B\)[/tex] is in the second quadrant. In the second quadrant, sine is positive and cosine is negative. We can find [tex]\(\sin(B)\)[/tex] and [tex]\(\cos(B)\)[/tex] using:
[tex]\[ \tan(B) = \frac{\sin(B)}{\cos(B)} \][/tex]
This implies:
[tex]\[ \sin(B) = -\sqrt{13} \cdot \cos(B) \][/tex]
Using the Pythagorean identity to find [tex]\(\cos(B)\)[/tex]:
[tex]\[ 1 + \tan^2(B) = \sec^2(B) \][/tex]
[tex]\[ 1 + (-\sqrt{13})^2 = \sec^2(B) \][/tex]
[tex]\[ 1 + 13 = \sec^2(B) \][/tex]
[tex]\[ 14 = \sec^2(B) \][/tex]
[tex]\[ \sec(B) = \pm \sqrt{14} \][/tex]
Since [tex]\(\frac{1}{\cos(B)} = \sec(B)\)[/tex], and [tex]\(\cos(B)\)[/tex] is negative in the second quadrant:
[tex]\[ \cos(B) = -\frac{1}{\sqrt{14}} \][/tex]
Using the identity:
[tex]\[ \sin(B) = \tan(B) \cdot \cos(B) \][/tex]
[tex]\[ \sin(B) = -\sqrt{13} \cdot -\frac{1}{\sqrt{14}} \][/tex]
[tex]\[ \sin(B) = \frac{\sqrt{13}}{\sqrt{14}} \][/tex]
### Step 3: Finding [tex]\(\tan(A - B)\)[/tex]
Using the tangent subtraction formula:
[tex]\[ \tan(A - B) = \frac{\tan(A) - \tan(B)}{1 + \tan(A) \cdot \tan(B)} \][/tex]
Substitute [tex]\(\tan(A)\)[/tex] and [tex]\(\tan(B)\)[/tex]:
[tex]\[ \tan(A - B) = \frac{-\frac{5}{12} - (-\sqrt{13})}{1 + \left(-\frac{5}{12}\right)(-\sqrt{13})} \][/tex]
[tex]\[ \tan(A - B) = \frac{-\frac{5}{12} + \sqrt{13}}{1 + \frac{5\sqrt{13}}{12}} \][/tex]
Simplify the numerator and the denominator:
[tex]\[ \tan(A - B) = \frac{\sqrt{13} - \frac{5}{12}}{1 + \frac{5\sqrt{13}}{12}} \][/tex]
[tex]\[ \tan(A - B) = \frac{12\sqrt{13} - 5}{12}{\frac{12}{12}}{12 + 5\sqrt{13}} \][/tex]
[tex]\[ \tan(A - B) = \frac{12\sqrt{13} - 5}{12 + 5\sqrt{13}} \][/tex]
Thus, the correct answer is:
[tex]\[ \tan(A-B) = 1.274374775947988 = \frac{12\sqrt{13} - 5}{12 + 5\sqrt{13}} \][/tex]
### Step 1: Finding [tex]\(\tan(A)\)[/tex]
Given:
[tex]\[ \sin(A) = \frac{5}{13} \][/tex]
Since [tex]\( \frac{\pi}{2} < A < \pi \)[/tex], angle [tex]\(A\)[/tex] is in the second quadrant. In the second quadrant, cosine is negative. We use the Pythagorean identity:
[tex]\[ \cos^2(A) + \sin^2(A) = 1 \][/tex]
[tex]\[ \cos^2(A) = 1 - \sin^2(A) \][/tex]
[tex]\[ \cos^2(A) = 1 - \left(\frac{5}{13}\right)^2 \][/tex]
[tex]\[ \cos^2(A) = 1 - \frac{25}{169} \][/tex]
[tex]\[ \cos^2(A) = \frac{144}{169} \][/tex]
[tex]\[ \cos(A) = -\sqrt{\frac{144}{169}} \][/tex]
[tex]\[ \cos(A) = -\frac{12}{13} \][/tex]
Now, we can find [tex]\(\tan(A)\)[/tex]:
[tex]\[ \tan(A) = \frac{\sin(A)}{\cos(A)} \][/tex]
[tex]\[ \tan(A) = \frac{\frac{5}{13}}{-\frac{12}{13}} \][/tex]
[tex]\[ \tan(A) = -\frac{5}{12} \][/tex]
### Step 2: Finding [tex]\(\tan(B)\)[/tex]
Given:
[tex]\[ \tan(B) = -\sqrt{13} \][/tex]
Since [tex]\( \frac{\pi}{2} < B < \pi \)[/tex], angle [tex]\(B\)[/tex] is in the second quadrant. In the second quadrant, sine is positive and cosine is negative. We can find [tex]\(\sin(B)\)[/tex] and [tex]\(\cos(B)\)[/tex] using:
[tex]\[ \tan(B) = \frac{\sin(B)}{\cos(B)} \][/tex]
This implies:
[tex]\[ \sin(B) = -\sqrt{13} \cdot \cos(B) \][/tex]
Using the Pythagorean identity to find [tex]\(\cos(B)\)[/tex]:
[tex]\[ 1 + \tan^2(B) = \sec^2(B) \][/tex]
[tex]\[ 1 + (-\sqrt{13})^2 = \sec^2(B) \][/tex]
[tex]\[ 1 + 13 = \sec^2(B) \][/tex]
[tex]\[ 14 = \sec^2(B) \][/tex]
[tex]\[ \sec(B) = \pm \sqrt{14} \][/tex]
Since [tex]\(\frac{1}{\cos(B)} = \sec(B)\)[/tex], and [tex]\(\cos(B)\)[/tex] is negative in the second quadrant:
[tex]\[ \cos(B) = -\frac{1}{\sqrt{14}} \][/tex]
Using the identity:
[tex]\[ \sin(B) = \tan(B) \cdot \cos(B) \][/tex]
[tex]\[ \sin(B) = -\sqrt{13} \cdot -\frac{1}{\sqrt{14}} \][/tex]
[tex]\[ \sin(B) = \frac{\sqrt{13}}{\sqrt{14}} \][/tex]
### Step 3: Finding [tex]\(\tan(A - B)\)[/tex]
Using the tangent subtraction formula:
[tex]\[ \tan(A - B) = \frac{\tan(A) - \tan(B)}{1 + \tan(A) \cdot \tan(B)} \][/tex]
Substitute [tex]\(\tan(A)\)[/tex] and [tex]\(\tan(B)\)[/tex]:
[tex]\[ \tan(A - B) = \frac{-\frac{5}{12} - (-\sqrt{13})}{1 + \left(-\frac{5}{12}\right)(-\sqrt{13})} \][/tex]
[tex]\[ \tan(A - B) = \frac{-\frac{5}{12} + \sqrt{13}}{1 + \frac{5\sqrt{13}}{12}} \][/tex]
Simplify the numerator and the denominator:
[tex]\[ \tan(A - B) = \frac{\sqrt{13} - \frac{5}{12}}{1 + \frac{5\sqrt{13}}{12}} \][/tex]
[tex]\[ \tan(A - B) = \frac{12\sqrt{13} - 5}{12}{\frac{12}{12}}{12 + 5\sqrt{13}} \][/tex]
[tex]\[ \tan(A - B) = \frac{12\sqrt{13} - 5}{12 + 5\sqrt{13}} \][/tex]
Thus, the correct answer is:
[tex]\[ \tan(A-B) = 1.274374775947988 = \frac{12\sqrt{13} - 5}{12 + 5\sqrt{13}} \][/tex]
