If `( sec A + tan A ) ( sec B + tan B ) ( sec C + tan C )=
( sec A - tan A ) ( sec B - tan B ) ( sec C - tan C )` then each side is equal to

Options

A. 1
B. `-1`
C. 0
D. none

Answer

To solve the problem, we need to analyze the given inequality:

\[ ( \sec A + \tan A )( \sec B + \tan B )( \sec C + \tan C ) \leq ( \sec A - \tan A )( \sec B - \tan B )( \sec C - \tan C ) \]

We will simplify both sides and find the equality.

### Step 1: Rewrite the expressions
Recall the identities:
\[ \sec A = \frac{1}{\cos A}, \quad \tan A = \frac{\sin A}{\cos A} \]
Thus,
\[ \sec A + \tan A = \frac{1 + \sin A}{\cos A}, \quad \sec A - \tan A = \frac{1 - \sin A}{\cos A} \]

### Step 2: Substitute into the inequality
Substituting these identities into the left-hand side (LHS) and right-hand side (RHS):
\[ LHS = \left( \frac{1 + \sin A}{\cos A} \right) \left( \frac{1 + \sin B}{\cos B} \right) \left( \frac{1 + \sin C}{\cos C} \right) \]
\[ RHS = \left( \frac{1 - \sin A}{\cos A} \right) \left( \frac{1 - \sin B}{\cos B} \right) \left( \frac{1 - \sin C}{\cos C} \right) \]

### Step 3: Simplify both sides
Now, we can simplify both sides:
\[ LHS = \frac{(1 + \sin A)(1 + \sin B)(1 + \sin C)}{\cos A \cos B \cos C} \]
\[ RHS = \frac{(1 - \sin A)(1 - \sin B)(1 - \sin C)}{\cos A \cos B \cos C} \]

### Step 4: Cancel out the common denominator
Since \(\cos A \cos B \cos C\) is common in both sides, we can cancel it out (assuming it is not zero):
\[ (1 + \sin A)(1 + \sin B)(1 + \sin C) \leq (1 - \sin A)(1 - \sin B)(1 - \sin C) \]

### Step 5: Analyze the inequality
This inequality can be analyzed further, but we can also consider the case when \(A = B = C = 0\):
\[ (1 + 0)(1 + 0)(1 + 0) = 1 \quad \text{and} \quad (1 - 0)(1 - 0)(1 - 0) = 1 \]
Thus, both sides are equal to 1.

### Step 6: Conclusion
Since the inequality holds true for \(A = B = C = 0\), we can conclude that the maximum value for both sides is indeed equal to 1.

Thus, the final answer is:
\[ \text{Each side is equal to } 1. \]