Find minimum value of `32cot^(2)theta+2tan^(2)theta`.

Answer

To find the minimum value of the expression \( 32 \cot^2 \theta + 2 \tan^2 \theta \), we can use the method of inequalities, specifically the Arithmetic Mean-Geometric Mean (AM-GM) inequality. Here’s a step-by-step solution:

### Step 1: Rewrite the expression
We start with the expression:
\[ y = 32 \cot^2 \theta + 2 \tan^2 \theta \]

### Step 2: Use the AM-GM Inequality
According to the AM-GM inequality, for any non-negative numbers \( a \) and \( b \):
\[ \frac{a + b}{2} \geq \sqrt{ab} \]
We can apply this to our two terms \( 32 \cot^2 \theta \) and \( 2 \tan^2 \theta \).

Let:
- \( a = 32 \cot^2 \theta \)
- \( b = 2 \tan^2 \theta \)

Then we have:
\[ \frac{32 \cot^2 \theta + 2 \tan^2 \theta}{2} \geq \sqrt{32 \cot^2 \theta \cdot 2 \tan^2 \theta} \]

### Step 3: Simplify the right-hand side
Now, we simplify the right-hand side:
\[ \sqrt{32 \cot^2 \theta \cdot 2 \tan^2 \theta} = \sqrt{64 \cot^2 \theta \tan^2 \theta} \]
Since \( \cot \theta = \frac{1}{\tan \theta} \), we have:
\[ \cot^2 \theta \tan^2 \theta = 1 \]
Thus,
\[ \sqrt{64 \cdot 1} = \sqrt{64} = 8 \]

### Step 4: Set up the inequality
Substituting back into our AM-GM inequality gives:
\[ \frac{32 \cot^2 \theta + 2 \tan^2 \theta}{2} \geq 8 \]
Multiplying both sides by 2:
\[ 32 \cot^2 \theta + 2 \tan^2 \theta \geq 16 \]

### Step 5: Conclusion
The minimum value of \( 32 \cot^2 \theta + 2 \tan^2 \theta \) is therefore:
\[ \boxed{16} \]