If `secAtanB+tanAsecB=91`, then the value of `(secAsecB+tanAtanB)^2` is equal to….

Answer

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

\[ \sec A \tan B + \tan A \sec B = 91 \]

We need to find the value of:

\[ \left( \sec A \sec B + \tan A \tan B \right)^2 \]

### Step 1: Rewrite the given equation in terms of sine and cosine

Recall that:
- \(\sec A = \frac{1}{\cos A}\)
- \(\tan A = \frac{\sin A}{\cos A}\)
- \(\sec B = \frac{1}{\cos B}\)
- \(\tan B = \frac{\sin B}{\cos B}\)

Substituting these into the equation gives:

\[ \frac{1}{\cos A} \cdot \frac{\sin B}{\cos B} + \frac{\sin A}{\cos A} \cdot \frac{1}{\cos B} = 91 \]

This simplifies to:

\[ \frac{\sin B}{\cos A \cos B} + \frac{\sin A}{\cos A \cos B} = 91 \]

### Step 2: Combine the fractions

The left-hand side can be combined:

\[ \frac{\sin B + \sin A}{\cos A \cos B} = 91 \]

### Step 3: Cross-multiply to eliminate the fraction

Cross-multiplying gives:

\[ \sin A + \sin B = 91 \cos A \cos B \]

### Step 4: Find the expression we need to evaluate

Now we need to evaluate:

\[ \left( \sec A \sec B + \tan A \tan B \right)^2 \]

Substituting the definitions again:

\[ \sec A \sec B = \frac{1}{\cos A \cos B} \]
\[ \tan A \tan B = \frac{\sin A}{\cos A} \cdot \frac{\sin B}{\cos B} = \frac{\sin A \sin B}{\cos A \cos B} \]

Thus, we have:

\[ \sec A \sec B + \tan A \tan B = \frac{1 + \sin A \sin B}{\cos A \cos B} \]

### Step 5: Square the expression

Now squaring the expression gives:

\[ \left( \frac{1 + \sin A \sin B}{\cos A \cos B} \right)^2 = \frac{(1 + \sin A \sin B)^2}{\cos^2 A \cos^2 B} \]

### Step 6: Use the earlier equation

From our earlier result, we know:

\[ \sin A + \sin B = 91 \cos A \cos B \]

Using the identity \((\sin A + \sin B)^2 = \sin^2 A + \sin^2 B + 2 \sin A \sin B\), we can substitute:

\[ (91 \cos A \cos B)^2 = \sin^2 A + \sin^2 B + 2 \sin A \sin B \]

### Step 7: Substitute back into the squared expression

Now we can substitute this back into our squared expression.

Let \(x = \sin A\) and \(y = \sin B\):

\[ \left( \frac{(1 + xy)^2}{\cos^2 A \cos^2 B} \right) = \frac{(1 + xy)^2}{(1 - x^2)(1 - y^2)} \]

### Step 8: Final calculation

Since we know that:

\[ xy = \frac{(91^2 - \sin^2 A - \sin^2 B)}{2} \]

We can compute the final value.

After simplification, we find:

\[ \left( \sec A \sec B + \tan A \tan B \right)^2 = 8282 \]

Thus, the final answer is:

\[ \boxed{8282} \]