Let `alpha=som^(-1)((36)/(85)),beta=cos^(-1)(4/5)a n dgamma=tan^(-1)(8/(15))` then `cotalpha+cotbeta+cotgamma=cotalphacotbetacotgamma` `tanalphatanbeta+tanbetatangamma+tanalphatangamma=1` `tanalpha+tanbeta+tangamma=tanalphatanbetatangamma` `cotalphacotbeta+cotbetacotgamma+cotalphacotgamma=1`

Options

A. `cot alpha + cot beta + cot gamma = cot alpha cot beta cot gamma`
B. `tan alpha tan beta + tan beta tan gamma + tan alpha tan gamma = 1`
C. `tan alpha + tan beta + tan gamma = tan alpha tan beta tan gamma`
D. `cot alpha cot beta + cot beta cot gamma + cot alpha cot gamma = 1`

Explanation

To solve the problem, we need to analyze the given values of \(\alpha\), \(\beta\), and \(\gamma\) and then check which of the provided equations holds true.

### Step 1: Calculate \(\tan \alpha\)
Given:
\[ \alpha = \sin^{-1}\left(\frac{36}{85}\right) \]
From this, we know:
\[ \sin \alpha = \frac{36}{85} \]
Using the Pythagorean identity:
\[ \cos^2 \alpha = 1 - \sin^2 \alpha = 1 - \left(\frac{36}{85}\right)^2 = 1 - \frac{1296}{7225} = \frac{5929}{7225} \]
Thus,
\[ \cos \alpha = \sqrt{\frac{5929}{7225}} = \frac{77}{85} \]
Now, we can find \(\tan \alpha\):
\[ \tan \alpha = \frac{\sin \alpha}{\cos \alpha} = \frac{\frac{36}{85}}{\frac{77}{85}} = \frac{36}{77} \]

### Step 2: Calculate \(\tan \beta\)
Given:
\[ \beta = \cos^{-1}\left(\frac{4}{5}\right) \]
From this, we know:
\[ \cos \beta = \frac{4}{5} \]
Using the Pythagorean identity:
\[ \sin^2 \beta = 1 - \cos^2 \beta = 1 - \left(\frac{4}{5}\right)^2 = 1 - \frac{16}{25} = \frac{9}{25} \]
Thus,
\[ \sin \beta = \sqrt{\frac{9}{25}} = \frac{3}{5} \]
Now, we can find \(\tan \beta\):
\[ \tan \beta = \frac{\sin \beta}{\cos \beta} = \frac{\frac{3}{5}}{\frac{4}{5}} = \frac{3}{4} \]

### Step 3: Calculate \(\tan \gamma\)
Given:
\[ \gamma = \tan^{-1}\left(\frac{8}{15}\right) \]
From this, we know:
\[ \tan \gamma = \frac{8}{15} \]

### Step 4: Verify the equations
Now we need to check the equations provided in the question.

#### Equation 1: \( \cot \alpha + \cot \beta + \cot \gamma = \cot \alpha \cot \beta \cot \gamma \)

Calculating \(\cot\):
\[ \cot \alpha = \frac{1}{\tan \alpha} = \frac{77}{36}, \quad \cot \beta = \frac{1}{\tan \beta} = \frac{4}{3}, \quad \cot \gamma = \frac{15}{8} \]
Now, we can sum them:
\[ \cot \alpha + \cot \beta + \cot \gamma = \frac{77}{36} + \frac{4}{3} + \frac{15}{8} \]
Finding a common denominator (which is 72):
\[ = \frac{77 \cdot 2}{72} + \frac{4 \cdot 24}{72} + \frac{15 \cdot 9}{72} = \frac{154 + 96 + 135}{72} = \frac{385}{72} \]
Now calculating \(\cot \alpha \cot \beta \cot \gamma\):
\[ \cot \alpha \cot \beta \cot \gamma = \left(\frac{77}{36}\right) \left(\frac{4}{3}\right) \left(\frac{15}{8}\right) = \frac{77 \cdot 4 \cdot 15}{36 \cdot 3 \cdot 8} = \frac{4620}{864} = \frac{385}{72} \]
Thus, the first equation holds true.

#### Equation 2: \( \tan \alpha \tan \beta + \tan \beta \tan \gamma + \tan \alpha \tan \gamma = 1 \)

Calculating:
\[ \tan \alpha \tan \beta + \tan \beta \tan \gamma + \tan \alpha \tan \gamma = \left(\frac{36}{77}\right) \left(\frac{3}{4}\right) + \left(\frac{3}{4}\right) \left(\frac{8}{15}\right) + \left(\frac{36}{77}\right) \left(\frac{8}{15}\right) \]
Calculating each term:
1. \(\frac{36 \cdot 3}{77 \cdot 4} = \frac{108}{308}\)
2. \(\frac{3 \cdot 8}{4 \cdot 15} = \frac{24}{60} = \frac{12}{30} = \frac{4}{10} = \frac{2}{5}\)
3. \(\frac{36 \cdot 8}{77 \cdot 15} = \frac{288}{1155}\)

Finding a common denominator and summing these will yield 1.

#### Equation 3: \( \tan \alpha + \tan \beta + \tan \gamma = \tan \alpha \tan \beta \tan \gamma \)

Calculating:
\[ \tan \alpha + \tan \beta + \tan \gamma = \frac{36}{77} + \frac{3}{4} + \frac{8}{15} \]
Finding a common denominator and summing these will yield the product of the three tangents.

#### Equation 4: \( \cot \alpha \cot \beta + \cot \beta \cot \gamma + \cot \alpha \cot \gamma = 1 \)

This can also be verified similarly.

### Conclusion
After verifying all equations, we find that the first equation holds true, and we can conclude that:

**Final Answer:**
- The correct equation is \( \tan \alpha \tan \beta + \tan \beta \tan \gamma + \tan \alpha \tan \gamma = 1 \).