If A and B are acute angles and SecA = 3,CotB = 4, then the value of `("cosec"^(2)A + sin^(2)B)/(cot^(2)A + sec^(2)B)` is:

Options

A. 2
B. `322/323`
C. `1/261`
D. `25/261`

Answer

To solve the problem, we need to find the value of \(\frac{\csc^2 A + \sin^2 B}{\cot^2 A + \sec^2 B}\) given that \(\sec A = 3\) and \(\cot B = 4\).

### Step-by-Step Solution:

1. **Find \(\csc^2 A\)**:
- We know that \(\sec A = 3\).
- Therefore, \(\cos A = \frac{1}{\sec A} = \frac{1}{3}\).
- Using the Pythagorean identity, \(\sin^2 A + \cos^2 A = 1\):
\[ \sin^2 A = 1 - \cos^2 A = 1 - \left(\frac{1}{3}\right)^2 = 1 - \frac{1}{9} = \frac{8}{9} \]
- Thus, \(\csc^2 A = \frac{1}{\sin^2 A} = \frac{9}{8}\).

2. **Find \(\sin^2 B\)**:
- We know that \(\cot B = 4\).
- Therefore, \(\tan B = \frac{1}{\cot B} = \frac{1}{4}\).
- Using the identity \(\sin^2 B + \cos^2 B = 1\) and knowing \(\tan B = \frac{\sin B}{\cos B}\):
- Let \(\sin B = x\) and \(\cos B = 4x\).
- Then, \(x^2 + (4x)^2 = 1\):
\[ x^2 + 16x^2 = 1 \implies 17x^2 = 1 \implies x^2 = \frac{1}{17} \implies \sin^2 B = \frac{1}{17} \]

3. **Find \(\cot^2 A\)**:
- We have \(\cos A = \frac{1}{3}\) and \(\sin A = \sqrt{\frac{8}{9}} = \frac{2\sqrt{2}}{3}\).
- Thus, \(\cot A = \frac{\cos A}{\sin A} = \frac{\frac{1}{3}}{\frac{2\sqrt{2}}{3}} = \frac{1}{2\sqrt{2}}\).
- Therefore, \(\cot^2 A = \left(\frac{1}{2\sqrt{2}}\right)^2 = \frac{1}{8}\).

4. **Find \(\sec^2 B\)**:
- We know \(\cot B = 4\), hence \(\tan B = \frac{1}{4}\).
- Using the identity \(\sec^2 B = 1 + \tan^2 B\):
\[ \sec^2 B = 1 + \left(\frac{1}{4}\right)^2 = 1 + \frac{1}{16} = \frac{17}{16} \]

5. **Substitute values into the expression**:
- Now we can substitute into the original expression:
\[ \frac{\csc^2 A + \sin^2 B}{\cot^2 A + \sec^2 B} = \frac{\frac{9}{8} + \frac{1}{17}}{\frac{1}{8} + \frac{17}{16}} \]

6. **Calculate the numerator**:
- Find a common denominator for the numerator:
\[ \text{LCM of } 8 \text{ and } 17 = 136 \]
\[ \frac{9}{8} = \frac{9 \times 17}{136} = \frac{153}{136}, \quad \frac{1}{17} = \frac{8}{136} \]
\[ \text{Numerator: } \frac{153 + 8}{136} = \frac{161}{136} \]

7. **Calculate the denominator**:
- Find a common denominator for the denominator:
\[ \text{LCM of } 8 \text{ and } 16 = 16 \]
\[ \frac{1}{8} = \frac{2}{16}, \quad \frac{17}{16} = \frac{17}{16} \]
\[ \text{Denominator: } \frac{2 + 17}{16} = \frac{19}{16} \]

8. **Final calculation**:
- Now substitute back into the expression:
\[ \frac{\frac{161}{136}}{\frac{19}{16}} = \frac{161}{136} \times \frac{16}{19} = \frac{161 \times 16}{136 \times 19} \]
- Simplifying gives:
\[ \frac{2576}{2584} = \frac{322}{323} \]

Thus, the final value is \(\frac{322}{323}\).