If `(secalpha+tanalpha)(secbeta+tanbeta)(secgamma+tangamma)=(secalpha-tanalpha)(secbeta-tanbeta)(secgamma-tangamma)`, then each of the side is equal to

Options

A. `pm1`
B. `-1`
C. `+1`
D. 4

Answer

To solve the equation

\[ (\sec \alpha + \tan \alpha)(\sec \beta + \tan \beta)(\sec \gamma + \tan \gamma) = (\sec \alpha - \tan \alpha)(\sec \beta - \tan \beta)(\sec \gamma - \tan \gamma), \]

we can start by simplifying the left-hand side and the right-hand side.

### Step 1: Rewrite the equation
Let:
\[ x = (\sec \alpha + \tan \alpha)(\sec \beta + \tan \beta)(\sec \gamma + \tan \gamma) \]
and
\[ y = (\sec \alpha - \tan \alpha)(\sec \beta - \tan \beta)(\sec \gamma - \tan \gamma). \]
We have \( x = y \).

### Step 2: Use the identity
Recall the identity:
\[ \sec^2 \theta - \tan^2 \theta = 1. \]
This means:
\[ (\sec \theta + \tan \theta)(\sec \theta - \tan \theta) = 1. \]

### Step 3: Apply the identity to each term
Applying this identity to each pair:
\[ (\sec \alpha + \tan \alpha)(\sec \alpha - \tan \alpha) = 1, \]
\[ (\sec \beta + \tan \beta)(\sec \beta - \tan \beta) = 1, \]
\[ (\sec \gamma + \tan \gamma)(\sec \gamma - \tan \gamma) = 1. \]

### Step 4: Multiply the results
Thus, we can write:
\[ x \cdot y = 1 \cdot 1 \cdot 1 = 1. \]

### Step 5: Set up the equation
Since \( x = y \), we can substitute:
\[ x^2 = 1. \]

### Step 6: Solve for \( x \)
Taking the square root of both sides gives:
\[ x = \pm 1. \]

### Conclusion
Thus, the value of each side (i.e., \( x \)) is either \( 1 \) or \( -1 \).