If `A+B=pi/3` where `A, B > 0,` then minimum value of `sec A + sec B` is equal to

Options

A. `4/sqrt3`
B. `8/sqrt3`
C. 6
D. None of these

Answer

To find the minimum value of \( \sec A + \sec B \) given that \( A + B = \frac{\pi}{3} \) (where \( A, B > 0 \)), we can use the method of Lagrange multipliers or apply the Arithmetic Mean-Geometric Mean (AM-GM) inequality. Here, we'll use the AM-GM inequality for simplicity.

### Step 1: Express \( \sec A + \sec B \)

We know that:
\[ \sec A = \frac{1}{\cos A}, \quad \sec B = \frac{1}{\cos B} \]
Thus, we need to minimize:
\[ \sec A + \sec B = \frac{1}{\cos A} + \frac{1}{\cos B} \]

### Step 2: Use the condition \( A + B = \frac{\pi}{3} \)

From the condition \( A + B = \frac{\pi}{3} \), we can express \( B \) in terms of \( A \):
\[ B = \frac{\pi}{3} - A \]

### Step 3: Rewrite the expression in terms of \( A \)

Substituting \( B \) into the expression for \( \sec A + \sec B \):
\[ \sec A + \sec B = \sec A + \sec\left(\frac{\pi}{3} - A\right) \]

### Step 4: Use the identity for secant

Using the identity \( \sec(\frac{\pi}{3} - A) = \frac{2}{\sqrt{3}} \sec A \) (derived from trigonometric identities), we can rewrite the expression:
\[ \sec\left(\frac{\pi}{3} - A\right) = \frac{2}{\sqrt{3}} \sec A \]

### Step 5: Apply AM-GM inequality

Now, we apply the AM-GM inequality:
\[ \sec A + \sec B \geq 2\sqrt{\sec A \sec B} \]
To find the minimum, we need to set \( A = B \). Given \( A + B = \frac{\pi}{3} \), we have:
\[ A = B = \frac{\pi}{6} \]

### Step 6: Calculate the minimum value

Now, substituting \( A = B = \frac{\pi}{6} \):
\[ \sec\left(\frac{\pi}{6}\right) = \frac{1}{\cos\left(\frac{\pi}{6}\right)} = \frac{1}{\frac{\sqrt{3}}{2}} = \frac{2}{\sqrt{3}} \]
Thus,
\[ \sec A + \sec B = 2 \cdot \sec\left(\frac{\pi}{6}\right) = 2 \cdot \frac{2}{\sqrt{3}} = \frac{4}{\sqrt{3}} \]

### Conclusion

The minimum value of \( \sec A + \sec B \) is:
\[ \frac{4}{\sqrt{3}} \]