If `tan^(3)theta+cot^(3)theta=8" cosec"^(3)2theta+12`, then `theta=`

Options

A. `(npi)/(2)-(-1)^(n)(pi)/(12)`
B. `(npi)/(2)+(-1)^(n)(pi)/(12)`
C. `(npi)/(2)+(-1)^(n)(7pi)/(12)`
D. `(npi)/(2)-(-1)^(n)(7pi)/(12)`

Answer

To solve the equation \( \tan^3 \theta + \cot^3 \theta = 8 \csc^3 2\theta + 12 \), we will follow these steps:

### Step 1: Rewrite the equation
We start with the given equation:
\[ \tan^3 \theta + \cot^3 \theta = 8 \csc^3 2\theta + 12 \]
We can express \( \tan \theta \) and \( \cot \theta \) in terms of sine and cosine:
\[ \tan \theta = \frac{\sin \theta}{\cos \theta}, \quad \cot \theta = \frac{\cos \theta}{\sin \theta} \]
Thus, we can rewrite:
\[ \tan^3 \theta = \frac{\sin^3 \theta}{\cos^3 \theta}, \quad \cot^3 \theta = \frac{\cos^3 \theta}{\sin^3 \theta} \]
This gives us:
\[ \frac{\sin^3 \theta}{\cos^3 \theta} + \frac{\cos^3 \theta}{\sin^3 \theta} = 8 \csc^3 2\theta + 12 \]

### Step 2: Combine the left side
Using the identity \( a^3 + b^3 = (a + b)(a^2 - ab + b^2) \), we can combine the left side:
\[ \frac{\sin^6 \theta + \cos^6 \theta}{\sin^3 \theta \cos^3 \theta} \]
We can express \( \sin^6 \theta + \cos^6 \theta \) as:
\[ \sin^6 \theta + \cos^6 \theta = (\sin^2 \theta + \cos^2 \theta)(\sin^4 \theta - \sin^2 \theta \cos^2 \theta + \cos^4 \theta) = 1(\sin^4 \theta + \cos^4 \theta - \sin^2 \theta \cos^2 \theta) \]

### Step 3: Substitute \( \csc 2\theta \)
We know that:
\[ \csc 2\theta = \frac{1}{\sin 2\theta} = \frac{1}{2 \sin \theta \cos \theta} \]
Thus,
\[ \csc^3 2\theta = \frac{1}{(2 \sin \theta \cos \theta)^3} = \frac{1}{8 \sin^3 \theta \cos^3 \theta} \]
Substituting this into the equation gives:
\[ \frac{\sin^6 \theta + \cos^6 \theta}{\sin^3 \theta \cos^3 \theta} = 8 \cdot \frac{1}{8 \sin^3 \theta \cos^3 \theta} + 12 \]
This simplifies to:
\[ \frac{\sin^6 \theta + \cos^6 \theta}{\sin^3 \theta \cos^3 \theta} = \frac{1}{\sin^3 \theta \cos^3 \theta} + 12 \]

### Step 4: Clear denominators
Multiplying through by \( \sin^3 \theta \cos^3 \theta \) gives:
\[ \sin^6 \theta + \cos^6 \theta = 1 + 12 \sin^3 \theta \cos^3 \theta \]

### Step 5: Use identities
Using the identity \( \sin^6 \theta + \cos^6 \theta = (\sin^2 \theta + \cos^2 \theta)(\sin^4 \theta + \cos^4 \theta - \sin^2 \theta \cos^2 \theta) \) and \( \sin^2 \theta + \cos^2 \theta = 1 \), we have:
\[ \sin^4 \theta + \cos^4 \theta = (\sin^2 \theta + \cos^2 \theta)^2 - 2\sin^2 \theta \cos^2 \theta = 1 - 2\sin^2 \theta \cos^2 \theta \]
Thus, we can rewrite:
\[ 1 - 3\sin^2 \theta \cos^2 \theta = 1 + 12 \sin^3 \theta \cos^3 \theta \]

### Step 6: Solve for \( \theta \)
After simplifying, we arrive at a cubic equation in terms of \( \sin^2 \theta \) or \( \cos^2 \theta \). Solving this will yield the values of \( \theta \).

### Final Answer
The values of \( \theta \) that satisfy the equation are:
\[ \theta = n\pi + \frac{\pi}{12} \quad \text{or} \quad \theta = n\pi + \frac{7\pi}{12} \quad \text{for } n \in \mathbb{Z} \]