If ` (1)/(cos alpha * cos beta )+ tan alpha * tan beta = tan gamma ` , where ` 0 lt gamma lt (pi)/2 and alpha , beta ` are positive acute angles , show that ` (pi)/4 lt gamma lt (pi)/2`

Answer

To solve the given problem, we start with the equation:

\[ \frac{1}{\cos \alpha \cos \beta} + \tan \alpha \tan \beta = \tan \gamma \]

where \(0 < \gamma < \frac{\pi}{2}\) and \(\alpha, \beta\) are positive acute angles. We need to show that:

\[ \frac{\pi}{4} < \gamma < \frac{\pi}{2} \]

### Step 1: Rewrite the equation

We can rewrite the equation by expressing \(\tan \alpha\) and \(\tan \beta\) in terms of \(\sec\):

\[ \frac{1}{\cos \alpha \cos \beta} + \tan \alpha \tan \beta = \tan \gamma \]

This can be rewritten as:

\[ \sec \alpha \sec \beta + \tan \alpha \tan \beta = \tan \gamma \]

### Step 2: Use the identity for \(\sec\) and \(\tan\)

Using the identity \(\sec^2 \theta = 1 + \tan^2 \theta\), we can express \(\sec \alpha\) and \(\sec \beta\):

\[ \sec \alpha = \sqrt{1 + \tan^2 \alpha}, \quad \sec \beta = \sqrt{1 + \tan^2 \beta} \]

### Step 3: Square both sides

Next, we square both sides of the equation:

\[ \left(\sec \alpha \sec \beta + \tan \alpha \tan \beta\right)^2 = \tan^2 \gamma \]

Expanding the left-hand side gives:

\[ \sec^2 \alpha \sec^2 \beta + 2 \sec \alpha \sec \beta \tan \alpha \tan \beta + \tan^2 \alpha \tan^2 \beta = \tan^2 \gamma \]

### Step 4: Substitute the identities

Substituting the identities we have:

\[ (1 + \tan^2 \alpha)(1 + \tan^2 \beta) + 2 \sec \alpha \sec \beta \tan \alpha \tan \beta + \tan^2 \alpha \tan^2 \beta = \tan^2 \gamma \]

### Step 5: Simplify the equation

After simplification, we get:

\[ 1 + \tan^2 \alpha + \tan^2 \beta + \tan^2 \alpha \tan^2 \beta + 2 \sec \alpha \sec \beta \tan \alpha \tan \beta = \tan^2 \gamma \]

### Step 6: Analyze the inequality

Since \(\alpha\) and \(\beta\) are acute angles, we know:

\[ \tan \alpha > 0 \quad \text{and} \quad \tan \beta > 0 \]

This implies that:

\[ \tan^2 \alpha \tan^2 \beta > 0 \]

Thus, we can conclude that:

\[ \tan^2 \gamma > 1 \implies \tan \gamma > 1 \]

### Step 7: Conclude the range for \(\gamma\)

Since \(\tan \gamma > 1\), we know:

\[ \gamma > \frac{\pi}{4} \]

And since \(\gamma < \frac{\pi}{2}\) (as given), we have:

\[ \frac{\pi}{4} < \gamma < \frac{\pi}{2} \]

### Final Conclusion

Thus, we have shown that:

\[ \frac{\pi}{4} < \gamma < \frac{\pi}{2} \]