If `(secA-tanA)(secB-tanB)(secC+tanC)=(secA+tanA)(secB+tanB)(secC-tanC)` then each side is equal to

Options

A. 0
B. 1
C. `-1`
D. `pm1`

Answer

To solve the equation \((\sec A - \tan A)(\sec B - \tan B)(\sec C + \tan C) = (\sec A + \tan A)(\sec B + \tan B)(\sec C - \tan C)\), we will simplify both sides step by step.

### Step 1: Use the identity for secant and tangent
Recall the identity:
\[ \sec x - \tan x = \frac{1}{\sec x + \tan x} \]
This means we can rewrite \((\sec A - \tan A)\) and \((\sec B - \tan B)\) in terms of \((\sec A + \tan A)\) and \((\sec B + \tan B)\).

### Step 2: Rewrite the left-hand side
Using the identity, we can express:
\[ \sec A - \tan A = \frac{1}{\sec A + \tan A} \]
Thus, the left-hand side becomes:
\[ \frac{1}{\sec A + \tan A} \cdot \frac{1}{\sec B + \tan B} \cdot (\sec C + \tan C) \]

### Step 3: Rewrite the right-hand side
Similarly, for the right-hand side:
\[ \sec A + \tan A \cdot \sec B + \tan B \cdot (\sec C - \tan C) = (\sec A + \tan A)(\sec B + \tan B) \cdot \frac{1}{\sec C + \tan C} \]

### Step 4: Set both sides equal
Now we have:
\[ \frac{(\sec C + \tan C)}{(\sec A + \tan A)(\sec B + \tan B)} = \frac{(\sec A + \tan A)(\sec B + \tan B)}{(\sec C + \tan C)} \]

### Step 5: Cross-multiply
Cross-multiplying gives us:
\[ (\sec C + \tan C)^2 = (\sec A + \tan A)(\sec B + \tan B)(\sec C + \tan C) \]

### Step 6: Factor out \((\sec C + \tan C)\)
Assuming \((\sec C + \tan C) \neq 0\), we can divide both sides by \((\sec C + \tan C)\):
\[ \sec C + \tan C = \sec A + \tan A + \sec B + \tan B \]

### Step 7: Conclude the equality
From the above steps, we can conclude that:
\[ (\sec A - \tan A)(\sec B - \tan B)(\sec C + \tan C) = (\sec A + \tan A)(\sec B + \tan B)(\sec C - \tan C) = \pm 1 \]

Thus, each side is equal to \(\pm 1\).