If A and B are complementary angles, find the value of `sqrt((tanAtanB+cotAcotB)/(sinAsecB)-(sin^(2)B)/(cos^(2)A))`

Options

A. 5
B. `sqrt(2)`
C. 1
D. 3

Answer

To solve the problem, we need to evaluate the expression:

\[ \sqrt{\frac{\tan A \tan B + \cot A \cot B}{\sin A \sec B - \frac{\sin^2 B}{\cos^2 A}}} \]

Given that \( A \) and \( B \) are complementary angles, we have:

\[ A + B = 90^\circ \quad \Rightarrow \quad B = 90^\circ - A \]

### Step 1: Substitute \( B \) in the expression

Substituting \( B = 90^\circ - A \) into the expression, we have:

\[ \sqrt{\frac{\tan A \tan(90^\circ - A) + \cot A \cot(90^\circ - A)}{\sin A \sec(90^\circ - A) - \frac{\sin^2(90^\circ - A)}{\cos^2 A}}} \]

### Step 2: Simplify \( \tan(90^\circ - A) \) and \( \cot(90^\circ - A) \)

Using the identities:
- \( \tan(90^\circ - A) = \cot A \)
- \( \cot(90^\circ - A) = \tan A \)
- \( \sec(90^\circ - A) = \csc A \)
- \( \sin(90^\circ - A) = \cos A \)

We can rewrite the expression as:

\[ \sqrt{\frac{\tan A \cot A + \cot A \tan A}{\sin A \csc A - \frac{\cos^2 A}{\cos^2 A}}} \]

### Step 3: Evaluate \( \tan A \cot A \)

Since \( \tan A \cot A = 1 \):

\[ \sqrt{\frac{1 + 1}{\sin A \csc A - 1}} = \sqrt{\frac{2}{\sin A \cdot \frac{1}{\sin A} - 1}} = \sqrt{\frac{2}{1 - 1}} = \sqrt{\frac{2}{0}} \]

### Step 4: Evaluate the denominator

Since the denominator simplifies to zero, we need to reevaluate our expression.

### Step 5: Correct the denominator

The denominator simplifies to:

\[ \sin A \cdot \frac{1}{\sin A} - 1 = 1 - 1 = 0 \]

### Final Step: Conclusion

Since the denominator approaches zero, the expression diverges, and thus the value of the original expression is undefined.