If `tan^(3)theta+cot^(3)theta=52`, then the value of ` tan^(2)theta+cot^(2)theta` is equal to :

Options

A. 14
B. 15
C. 16
D. 17

Answer

To solve the problem, we start with the equation given:

\[ \tan^3 \theta + \cot^3 \theta = 52 \]

### Step 1: Use the identity for the sum of cubes
We can express \(\tan^3 \theta + \cot^3 \theta\) using the identity for the sum of cubes:

\[ a^3 + b^3 = (a + b)(a^2 - ab + b^2) \]

Let \(a = \tan \theta\) and \(b = \cot \theta\). Thus, we have:

\[ \tan^3 \theta + \cot^3 \theta = (\tan \theta + \cot \theta)((\tan^2 \theta + \cot^2 \theta) - \tan \theta \cot \theta) \]

### Step 2: Substitute \(\tan \theta \cot \theta\)
Since \(\tan \theta \cot \theta = 1\), we can rewrite the equation as:

\[ \tan^3 \theta + \cot^3 \theta = (\tan \theta + \cot \theta)(\tan^2 \theta + \cot^2 \theta - 1) \]

### Step 3: Set up the equation
Now substituting the given value:

\[ 52 = (\tan \theta + \cot \theta)(\tan^2 \theta + \cot^2 \theta - 1) \]

### Step 4: Let \(T = \tan \theta + \cot \theta\)
Let \(T = \tan \theta + \cot \theta\). We can express \(\tan^2 \theta + \cot^2 \theta\) in terms of \(T\):

\[ \tan^2 \theta + \cot^2 \theta = T^2 - 2 \]

Substituting this into our equation gives:

\[ 52 = T((T^2 - 2) - 1) \]

This simplifies to:

\[ 52 = T(T^2 - 3) \]

### Step 5: Rearranging the equation
Rearranging gives us:

\[ T^3 - 3T - 52 = 0 \]

### Step 6: Solve for \(T\) using trial and error
We can use trial and error to find \(T\). Testing \(T = 4\):

\[ 4^3 - 3(4) - 52 = 64 - 12 - 52 = 0 \]

Thus, \(T = 4\) is a solution.

### Step 7: Find \(\tan^2 \theta + \cot^2 \theta\)
Now substituting back to find \(\tan^2 \theta + \cot^2 \theta\):

\[ \tan^2 \theta + \cot^2 \theta = T^2 - 2 = 4^2 - 2 = 16 - 2 = 14 \]

### Final Answer
Thus, the value of \(\tan^2 \theta + \cot^2 \theta\) is:

\[ \boxed{14} \]