If `tan(11theta)=cot(7theta)`, then what is the value of `sin^2(6theta)+sec^2(9theta)+cosec^2(12theta)`?

Options

A. `35/12`
B. `23/6`
C. `31/12`
D. `43/12`

Answer

To solve the equation \( \tan(11\theta) = \cot(7\theta) \), we can start by using the identity that relates tangent and cotangent.

### Step 1: Rewrite the equation using trigonometric identities
We know that:
\[ \cot(x) = \frac{1}{\tan(x)} \]
Thus, we can rewrite the equation as:
\[ \tan(11\theta) = \frac{1}{\tan(7\theta)} \]
This implies:
\[ \tan(11\theta) \tan(7\theta) = 1 \]

### Step 2: Use the identity for tangent
From the identity \( \tan(A) \tan(B) = 1 \), we can conclude:
\[ 11\theta + 7\theta = 90^\circ \quad \text{(since } \tan(A) \tan(B) = 1 \text{ when } A + B = 90^\circ\text{)} \]
This simplifies to:
\[ 18\theta = 90^\circ \]

### Step 3: Solve for \( \theta \)
Dividing both sides by 18:
\[ \theta = \frac{90^\circ}{18} = 5^\circ \]

### Step 4: Calculate \( \sin^2(6\theta) + \sec^2(9\theta) + \csc^2(12\theta) \)
Now we will substitute \( \theta = 5^\circ \) into the expression \( \sin^2(6\theta) + \sec^2(9\theta) + \csc^2(12\theta) \).

1. **Calculate \( 6\theta \)**:
\[ 6\theta = 6 \times 5^\circ = 30^\circ \]
Thus:
\[ \sin^2(6\theta) = \sin^2(30^\circ) = \left(\frac{1}{2}\right)^2 = \frac{1}{4} \]

2. **Calculate \( 9\theta \)**:
\[ 9\theta = 9 \times 5^\circ = 45^\circ \]
Thus:
\[ \sec^2(9\theta) = \sec^2(45^\circ) = \left(\frac{1}{\cos(45^\circ)}\right)^2 = \left(\frac{1}{\frac{1}{\sqrt{2}}}\right)^2 = 2 \]

3. **Calculate \( 12\theta \)**:
\[ 12\theta = 12 \times 5^\circ = 60^\circ \]
Thus:
\[ \csc^2(12\theta) = \csc^2(60^\circ) = \left(\frac{1}{\sin(60^\circ)}\right)^2 = \left(\frac{1}{\frac{\sqrt{3}}{2}}\right)^2 = \frac{4}{3} \]

### Step 5: Combine the results
Now we can sum these values:
\[ \sin^2(6\theta) + \sec^2(9\theta) + \csc^2(12\theta) = \frac{1}{4} + 2 + \frac{4}{3} \]

### Step 6: Find a common denominator
The common denominator for \( 4, 1, \) and \( 3 \) is \( 12 \):
\[ \frac{1}{4} = \frac{3}{12}, \quad 2 = \frac{24}{12}, \quad \frac{4}{3} = \frac{16}{12} \]
Now, adding these fractions:
\[ \frac{3}{12} + \frac{24}{12} + \frac{16}{12} = \frac{43}{12} \]

### Final Answer
Thus, the value of \( \sin^2(6\theta) + \sec^2(9\theta) + \csc^2(12\theta) \) is:
\[ \boxed{\frac{43}{12}} \]