Let `vecV(theta)=(cos theta+sectheta), hata +(cos theta-sec theta)` where `hata` and `hatb` are unit vectors and the angle between `hata` and `vecg` is `60^(@)`, then the minimum value of `|vecV|^(4)` is equal to

Answer

To solve the problem, we need to find the minimum value of \( |\vec{V}|^4 \) where

\[ \vec{V}(\theta) = (\cos \theta + \sec \theta) \hat{a} + (\cos \theta - \sec \theta) \hat{b} \]

and the angle between the unit vectors \( \hat{a} \) and \( \hat{b} \) is \( 60^\circ \).

### Step 1: Calculate the Magnitude of \( \vec{V} \)

The magnitude of \( \vec{V} \) can be calculated using the formula for the magnitude of a vector:

\[ |\vec{V}| = \sqrt{(\cos \theta + \sec \theta)^2 + (\cos \theta - \sec \theta)^2 + 2(\cos \theta + \sec \theta)(\cos \theta - \sec \theta) \cos(60^\circ)} \]

### Step 2: Simplify the Expression

Using \( \cos(60^\circ) = \frac{1}{2} \):

\[ |\vec{V}| = \sqrt{(\cos \theta + \sec \theta)^2 + (\cos \theta - \sec \theta)^2 + (\cos \theta + \sec \theta)(\cos \theta - \sec \theta)} \]

### Step 3: Expand the Squares

Expanding the squares:

\[ (\cos \theta + \sec \theta)^2 = \cos^2 \theta + 2\cos \theta \sec \theta + \sec^2 \theta \]
\[ (\cos \theta - \sec \theta)^2 = \cos^2 \theta - 2\cos \theta \sec \theta + \sec^2 \theta \]

Adding these gives:

\[ 2\cos^2 \theta + 2\sec^2 \theta \]

### Step 4: Combine Terms

Now, combine the terms:

\[ |\vec{V}|^2 = 2\cos^2 \theta + 2\sec^2 \theta + (\cos^2 \theta - \sec^2 \theta) \]

This simplifies to:

\[ |\vec{V}|^2 = 3\cos^2 \theta + \sec^2 \theta \]

### Step 5: Substitute \( \sec^2 \theta \)

Recall that \( \sec^2 \theta = 1 + \tan^2 \theta \):

\[ |\vec{V}|^2 = 3\cos^2 \theta + 1 + \tan^2 \theta \]

### Step 6: Find Minimum Value

To find the minimum value of \( |\vec{V}|^4 \), we need to minimize \( |\vec{V}|^2 \):

Let \( x = \cos^2 \theta \), then \( \tan^2 \theta = \frac{1 - x}{x} \):

\[ |\vec{V}|^2 = 3x + 1 + \frac{1 - x}{x} \]

### Step 7: Differentiate and Solve

To find the minimum, differentiate and set the derivative to zero:

\[ \frac{d}{dx}(3x + 1 + \frac{1 - x}{x}) = 3 - \frac{1}{x^2} = 0 \]

Solving gives:

\[ 3x^2 - 1 = 0 \implies x^2 = \frac{1}{3} \implies x = \frac{1}{\sqrt{3}} \]

### Step 8: Calculate Minimum Value

Substituting back to find \( |\vec{V}|^2 \):

\[ |\vec{V}|^2 = 3 \cdot \frac{1}{3} + 1 + 1 = 2 \]

Thus,

\[ |\vec{V}|^4 = (|\vec{V}|^2)^2 = 2^2 = 4 \]

### Final Answer

The minimum value of \( |\vec{V}|^4 \) is:

\[ \boxed{4} \]