If `alpha+beta=90^(@)`,then the value of `(1-sin^(2)alpha)(1-cos^(2)alpha)xx(1+cot^(2)beta)(1+tan^(2)beta)`is

Options

A. 1
B. `-1`
C. 0
D. 2

Answer

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

\[ (1 - \sin^2 \alpha)(1 - \cos^2 \alpha)(1 + \cot^2 \beta)(1 + \tan^2 \beta) \]

Given that \(\alpha + \beta = 90^\circ\), we can use some trigonometric identities to simplify the expression.

### Step 1: Simplify \(1 - \sin^2 \alpha\) and \(1 - \cos^2 \alpha\)

Using the Pythagorean identity:

\[ 1 - \sin^2 \alpha = \cos^2 \alpha \]
\[ 1 - \cos^2 \alpha = \sin^2 \alpha \]

So, we can rewrite the expression as:

\[ \cos^2 \alpha \cdot \sin^2 \alpha \cdot (1 + \cot^2 \beta)(1 + \tan^2 \beta) \]

### Step 2: Simplify \(1 + \cot^2 \beta\) and \(1 + \tan^2 \beta\)

Using the identities:

\[ 1 + \cot^2 \beta = \frac{1}{\sin^2 \beta} \]
\[ 1 + \tan^2 \beta = \frac{1}{\cos^2 \beta} \]

Thus, we can rewrite the expression further:

\[ \cos^2 \alpha \cdot \sin^2 \alpha \cdot \frac{1}{\sin^2 \beta} \cdot \frac{1}{\cos^2 \beta} \]

### Step 3: Combine the terms

Now, we can combine the terms:

\[ \cos^2 \alpha \cdot \sin^2 \alpha \cdot \frac{1}{\sin^2 \beta \cdot \cos^2 \beta} \]

### Step 4: Use the identity for \(\alpha + \beta = 90^\circ\)

Since \(\beta = 90^\circ - \alpha\), we have:

\[ \sin \beta = \cos \alpha \quad \text{and} \quad \cos \beta = \sin \alpha \]

Substituting these into our expression gives:

\[ \cos^2 \alpha \cdot \sin^2 \alpha \cdot \frac{1}{\cos^2 \alpha \cdot \sin^2 \alpha} \]

### Step 5: Simplify the expression

This simplifies to:

\[ 1 \]

Thus, the final value of the expression is:

\[ \boxed{1} \]