Let `alpha,beta,gamma > 0 and alpha+beta+gamma=pi/2.` Statement-1: `|tan alpha tan beta-(a!)/6|+|tan beta tan gamma-(b!)/2|+|tan gamma tanalpha-(c!)/3| le 0,` where `n! =1.2..........n,` then `tan alpha tanbeta,tanbeta tangamma, tan gamma tan alpha=1`

Settlement 2 : `tan alpha tanbeta+,tanbeta tangamma+, tan gamma tan alpha=1`

Answer

To solve the given problem step by step, let's analyze the statements provided and derive the necessary conclusions.

### Step 1: Understanding the Given Conditions
We are given that:
- \( \alpha, \beta, \gamma > 0 \)
- \( \alpha + \beta + \gamma = \frac{\pi}{2} \)

### Step 2: Analyzing Statement 2
Statement 2 claims:
\[ \tan \alpha \tan \beta + \tan \beta \tan \gamma + \tan \gamma \tan \alpha = 1 \]

To prove this, we can use the identity for the tangent of a sum:
\[ \tan(\alpha + \beta) = \tan\left(\frac{\pi}{2} - \gamma\right) = \cot \gamma \]

Using the tangent addition formula:
\[ \tan(\alpha + \beta) = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta} \]

Setting this equal to \( \cot \gamma \):
\[ \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta} = \cot \gamma = \frac{1}{\tan \gamma} \]

Cross-multiplying gives:
\[ (\tan \alpha + \tan \beta) \tan \gamma = 1 - \tan \alpha \tan \beta \]

Rearranging this, we find:
\[ \tan \alpha \tan \beta + \tan \beta \tan \gamma + \tan \gamma \tan \alpha = 1 \]

Thus, Statement 2 is **true**.

### Step 3: Analyzing Statement 1
Statement 1 claims:
\[ |\tan \alpha \tan \beta - (a!)/6| + |\tan \beta \tan \gamma - (b!)/2| + |\tan \gamma \tan \alpha - (c!)/3| \leq 0 \]

Since the absolute value is always non-negative, the only way for the sum of these absolute values to be less than or equal to zero is if each term is exactly zero. Therefore, we can conclude:
\[ \tan \alpha \tan \beta = \frac{a!}{6}, \quad \tan \beta \tan \gamma = \frac{b!}{2}, \quad \tan \gamma \tan \alpha = \frac{c!}{3} \]

However, we have already established from Statement 2 that:
\[ \tan \alpha \tan \beta + \tan \beta \tan \gamma + \tan \gamma \tan \alpha = 1 \]

This contradicts the possibility of the absolute values being less than or equal to zero unless all terms are zero, which is impossible since \( \alpha, \beta, \gamma > 0 \). Therefore, Statement 1 is **false**.

### Conclusion
- Statement 1 is **false**.
- Statement 2 is **true**.