`(secx.secy + tanx.tany)^(2)-(secx.tany + tanx.secy)^(2)` in its simplest form, is

Options

A. a)`-1`
B. b)0
C. c)`sec^(2)x`
D. d)1

Answer

To simplify the expression \((\sec x \sec y + \tan x \tan y)^{2} - (\sec x \tan y + \tan x \sec y)^{2}\), we can follow these steps:

### Step 1: Identify the expressions
Let:
- \( a = \sec x \sec y \)
- \( b = \tan x \tan y \)
- \( c = \sec x \tan y \)
- \( d = \tan x \sec y \)

The expression can be rewritten as:
\[ (a + b)^{2} - (c + d)^{2} \]

### Step 2: Apply the difference of squares formula
Using the difference of squares formula, \( A^2 - B^2 = (A - B)(A + B) \), we can rewrite our expression:
\[ ((a + b) - (c + d))((a + b) + (c + d)) \]

### Step 3: Simplify the first part
Now, simplify \((a + b) - (c + d)\):
\[ (a + b) - (c + d) = \sec x \sec y + \tan x \tan y - (\sec x \tan y + \tan x \sec y) \]
This simplifies to:
\[ \sec x \sec y - \sec x \tan y + \tan x \tan y - \tan x \sec y \]
Rearranging gives:
\[ \sec x (\sec y - \tan y) + \tan x (\tan y - \sec y) \]

### Step 4: Simplify the second part
Now simplify \((a + b) + (c + d)\):
\[ (a + b) + (c + d) = \sec x \sec y + \tan x \tan y + \sec x \tan y + \tan x \sec y \]
This can be rearranged to:
\[ \sec x \sec y + \sec x \tan y + \tan x \sec y + \tan x \tan y \]

### Step 5: Factor out common terms
Now we can factor out common terms from both parts:
1. From \((a + b) - (c + d)\), we notice that \(\sec y - \tan y\) and \(\tan y - \sec y\) can be factored.
2. From \((a + b) + (c + d)\), we can factor out \(\sec x\) and \(\tan x\).

### Step 6: Use trigonometric identities
Using the identity \(\sec^2 \theta - \tan^2 \theta = 1\), we can simplify our expression further:
\[ \sec^2 x - \tan^2 x = 1 \]
Thus, we can conclude that:
\[ \sec^2 x \sec^2 y + \tan^2 x \tan^2 y - \sec^2 x \tan^2 y - \tan^2 x \sec^2 y = 1 \]

### Final Result
The expression simplifies to:
\[ 1 \]