The local maximum value of `f(x)=(sin^(2)x)/(sin^(2)x-sin^(2)a) (0 lt a lt pi//2)` is :

Options

A. 0
B. `sec^(2) a`
C. `cosec^(2) a`
D. None of these

Answer

To find the local maximum value of the function \( f(x) = \frac{\sin^2 x}{\sin^2 x - \sin^2 a} \) where \( 0 < a < \frac{\pi}{2} \), we will follow these steps:

### Step 1: Differentiate the Function
We will use the quotient rule to differentiate \( f(x) \). The quotient rule states that if \( f(x) = \frac{u(x)}{v(x)} \), then \( f'(x) = \frac{u'v - uv'}{v^2} \).

Here, let:
- \( u(x) = \sin^2 x \)
- \( v(x) = \sin^2 x - \sin^2 a \)

Now, we differentiate \( u(x) \) and \( v(x) \):
- \( u'(x) = 2 \sin x \cos x = \sin 2x \)
- \( v'(x) = 2 \sin x \cos x = \sin 2x \)

Now applying the quotient rule:
\[ f'(x) = \frac{(\sin 2x)(\sin^2 x - \sin^2 a) - (\sin^2 x)(\sin 2x)}{(\sin^2 x - \sin^2 a)^2} \]

### Step 2: Simplify the Derivative
The numerator simplifies as follows:
\[ f'(x) = \frac{\sin 2x (\sin^2 x - \sin^2 a - \sin^2 x)}{(\sin^2 x - \sin^2 a)^2} = \frac{-\sin 2x \sin^2 a}{(\sin^2 x - \sin^2 a)^2} \]

### Step 3: Set the Derivative to Zero
To find critical points, set \( f'(x) = 0 \):
\[ -\sin 2x \sin^2 a = 0 \]
This gives us:
\[ \sin 2x = 0 \]
The solutions for \( 2x = n\pi \) where \( n \) is an integer, leads to:
\[ x = \frac{n\pi}{2} \]

### Step 4: Determine Valid Critical Points
Since \( 0 < a < \frac{\pi}{2} \), we will consider \( n = 0 \) and \( n = 1 \):
- For \( n = 0 \): \( x = 0 \)
- For \( n = 1 \): \( x = \frac{\pi}{2} \)

### Step 5: Evaluate the Function at Critical Points
Evaluate \( f(x) \) at the critical points:
1. \( f(0) = \frac{\sin^2 0}{\sin^2 0 - \sin^2 a} = \frac{0}{- \sin^2 a} = 0 \)
2. \( f\left(\frac{\pi}{2}\right) = \frac{\sin^2 \frac{\pi}{2}}{\sin^2 \frac{\pi}{2} - \sin^2 a} = \frac{1}{1 - \sin^2 a} = \frac{1}{\cos^2 a} = \sec^2 a \)

### Step 6: Conclusion
The local maximum value of \( f(x) \) occurs at \( x = \frac{\pi}{2} \) and is given by:
\[ \sec^2 a \]

### Final Answer
The local maximum value of \( f(x) \) is \( \sec^2 a \).

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