To solve the problem, we need to find the values of angles B and C in triangle ABC using the given inverse trigonometric functions, and then determine the relationship between tan A, tan B, and tan C.
### Step 1: Calculate angle B
Given:
\[
\angle B = \sec^{-1}\left(\frac{5}{4}\right) + \csc^{-1}(\sqrt{5})
\]
Let:
- \( \sec A = \frac{5}{4} \)
- \( \csc B = \sqrt{5} \)
From \( \sec A = \frac{5}{4} \), we can form a right triangle:
- Hypotenuse = 5
- Base = 4
- Using Pythagorean theorem, the height (perpendicular) = \( \sqrt{5^2 - 4^2} = \sqrt{25 - 16} = 3 \)
Thus,
\[
\tan A = \frac{\text{opposite}}{\text{adjacent}} = \frac{3}{4}
\]
Now for \( \csc B = \sqrt{5} \):
- Hypotenuse = \( \sqrt{5} \)
- Using Pythagorean theorem, the base = 2 (since \( \sin B = \frac{1}{\sqrt{5}} \) gives us the opposite side as 1)
- Therefore, \( \tan B = \frac{1}{2} \)
Now, we can find \( \tan(\angle B) \) using the formula for the tangent of the sum of two angles:
\[
\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}
\]
Substituting the values:
\[
\tan B = \tan\left(\tan^{-1}\left(\frac{3}{4}\right) + \tan^{-1}\left(\frac{1}{2}\right)\right) = \frac{\frac{3}{4} + \frac{1}{2}}{1 - \frac{3}{4} \cdot \frac{1}{2}} = \frac{\frac{3}{4} + \frac{2}{4}}{1 - \frac{3}{8}} = \frac{\frac{5}{4}}{\frac{5}{8}} = 2
\]
Thus,
\[
\tan B = 2
\]
### Step 2: Calculate angle C
Given:
\[
\angle C = \csc^{-1}\left(\frac{25}{7}\right) + \cot^{-1}\left(\frac{9}{13}\right)
\]
Let:
- \( \csc \alpha = \frac{25}{7} \)
- \( \cot \beta = \frac{9}{13} \)
From \( \csc \alpha = \frac{25}{7} \):
- Hypotenuse = 25
- Opposite = 7
- Using Pythagorean theorem, the base = \( \sqrt{25^2 - 7^2} = \sqrt{625 - 49} = \sqrt{576} = 24 \)
Thus,
\[
\tan \alpha = \frac{7}{24}
\]
For \( \cot \beta = \frac{9}{13} \):
- Therefore, \( \tan \beta = \frac{1}{\cot \beta} = \frac{13}{9} \)
Now, we can find \( \tan(\angle C) \) using the formula for the tangent of the sum of two angles:
\[
\tan C = \tan\left(\tan^{-1}\left(\frac{7}{24}\right) + \tan^{-1}\left(\frac{13}{9}\right)\right) = \frac{\frac{7}{24} + \frac{13}{9}}{1 - \frac{7}{24} \cdot \frac{13}{9}}
\]
Calculating this gives:
\[
\tan C = \frac{\frac{63 + 104}{216}}{1 - \frac{91}{216}} = \frac{\frac{167}{216}}{\frac{125}{216}} = \frac{167}{125}
\]
### Step 3: Calculate angle A
Using the fact that the sum of angles in a triangle equals \( \pi \):
\[
\angle A + \angle B + \angle C = \pi
\]
Substituting the known angles:
\[
\angle A + \tan^{-1}(2) + \tan^{-1}(3) = \pi
\]
Using the tangent addition formula again:
\[
\tan A = \tan\left(\pi - \left(\tan^{-1}(2) + \tan^{-1}(3)\right)\right) = -\tan\left(\tan^{-1}(2) + \tan^{-1}(3)\right)
\]
Calculating gives:
\[
\tan A = -\frac{5}{-1} = -5
\]
### Step 4: Determine the relationship between \( \tan A, \tan B, \tan C \)
Now we have:
- \( \tan A = -5 \)
- \( \tan B = 2 \)
- \( \tan C = \frac{167}{125} \)
To check if they are in AP, GP, or HP:
- Check if \( 2 \) is the arithmetic mean of \( -5 \) and \( \frac{167}{125} \):
\[
2 = \frac{-5 + \frac{167}{125}}{2}
\]
This does not hold true.
- Check for geometric mean:
\[
2^2 = -5 \cdot \frac{167}{125}
\]
This does not hold true.
- Check for harmonic mean:
\[
\frac{2}{\frac{1}{-5} + \frac{1}{2} + \frac{1}{\frac{167}{125}}}
\]
This does not hold true.
Thus, \( \tan A, \tan B, \tan C \) are in **AP**.
