If `A+B=90^(@)`, then
`(tanAtanB+tanAcotB)/(sinAsecB)-(sin^(2)B)/(cos^(2)A)`
is equal to ____.

Options

A. `cot^(2)A`
B. `cot^(2)B`
C. `-tan^(2)A`
D. `-cot^(2)A`

Answer

To solve the expression \((\tan A \tan B + \tan A \cot B) / (\sin A \sec B) - \frac{\sin^2 B}{\cos^2 A}\) given that \(A + B = 90^\circ\), we can follow these steps:

### Step 1: Use the identity for angles
Since \(A + B = 90^\circ\), we can express \(A\) in terms of \(B\):
\[ A = 90^\circ - B \]

### Step 2: Substitute \(\tan A\) and \(\sin A\)
Using the co-function identities:
\[ \tan A = \tan(90^\circ - B) = \cot B \]
\[ \sin A = \sin(90^\circ - B) = \cos B \]

### Step 3: Substitute into the expression
Now substituting \(\tan A\) and \(\sin A\) into the original expression:
\[ \frac{\tan A \tan B + \tan A \cot B}{\sin A \sec B} - \frac{\sin^2 B}{\cos^2 A} \]
becomes:
\[ \frac{\cot B \tan B + \cot B \cot B}{\cos B \sec B} - \frac{\sin^2 B}{\cos^2(90^\circ - B)} \]

### Step 4: Simplify \(\tan B\) and \(\sec B\)
Recall that:
\[ \cot B = \frac{1}{\tan B} \quad \text{and} \quad \sec B = \frac{1}{\cos B} \]
Thus, we can simplify:
\[ \frac{\cot B \tan B + \cot^2 B}{\cos B \cdot \frac{1}{\cos B}} - \frac{\sin^2 B}{\sin^2 B} \]
This simplifies to:
\[ \cot B + \cot^2 B - 1 \]

### Step 5: Use the identity for cotangent
Using the identity \(\cot^2 B + 1 = \csc^2 B\), we can rewrite:
\[ \cot B + \cot^2 B - 1 = \cot B + (\csc^2 B - 1) - 1 \]
This simplifies to:
\[ \cot B + \csc^2 B - 2 \]

### Step 6: Final simplification
Since \(\csc^2 B = 1 + \cot^2 B\), we have:
\[ \cot B + (1 + \cot^2 B) - 2 = \cot B + \cot^2 B - 1 \]
Thus, the final expression simplifies to:
\[ \cot^2 B \]

### Conclusion
The final result is:
\[ \cot^2 B \]