The minimum value of `(a^(2))/(cos^(2)x)+(b^(2))/(sin^(2)x)`

Options

A. `(A-b)^(2)`
B. `a^(2)+b^(2)`
C. `(a+b)^(2)`
D. `a^(2)-b^(2)`

Answer

To find the minimum value of the expression \(\frac{a^2}{\cos^2 x} + \frac{b^2}{\sin^2 x}\), we can use the method of inequalities, specifically the Cauchy-Schwarz inequality.

### Step-by-Step Solution:

1. **Identify the Expression**: We need to minimize the expression:
\[ E = \frac{a^2}{\cos^2 x} + \frac{b^2}{\sin^2 x} \]

2. **Apply Cauchy-Schwarz Inequality**: According to the Cauchy-Schwarz inequality:
\[ (u_1^2 + u_2^2)(v_1^2 + v_2^2) \geq (u_1v_1 + u_2v_2)^2 \]
We can set \(u_1 = \frac{a}{\cos x}\), \(u_2 = \frac{b}{\sin x}\), \(v_1 = \cos x\), and \(v_2 = \sin x\).

3. **Substitute into Cauchy-Schwarz**:
\[ \left(\frac{a^2}{\cos^2 x} + \frac{b^2}{\sin^2 x}\right)(\cos^2 x + \sin^2 x) \geq (a + b)^2 \]
Since \(\cos^2 x + \sin^2 x = 1\), we have:
\[ \frac{a^2}{\cos^2 x} + \frac{b^2}{\sin^2 x} \geq (a + b)^2 \]

4. **Find Conditions for Equality**: The equality in the Cauchy-Schwarz inequality holds when:
\[ \frac{a}{\cos x} = \frac{b}{\sin x} \implies a \sin x = b \cos x \]
This can be rearranged to:
\[ \tan x = \frac{a}{b} \]

5. **Determine Minimum Value**: The minimum value of \(E\) occurs when \(\tan x = \frac{a}{b}\). Substituting this back into our expression:
\[ E_{\text{min}} = (a + b)^2 \]

6. **Final Result**: Therefore, the minimum value of \(\frac{a^2}{\cos^2 x} + \frac{b^2}{\sin^2 x}\) is:
\[ \boxed{4ab} \]