If `cos^(-1) . x/a - sin^(-1). y/b = theta (a , b , ne 0)`, then the maximum value of `b^(2) x^(2) + a^(2) y^(2) + 2ab xy sin theta ` equals

Options

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

Answer

To solve the problem, we need to find the maximum value of the expression \( b^2 x^2 + a^2 y^2 + 2ab xy \sin \theta \) given the equation \( \cos^{-1} \left( \frac{x}{a} \right) - \sin^{-1} \left( \frac{y}{b} \right) = \theta \).

### Step 1: Rewrite the equation
We start with the equation:
\[ \cos^{-1} \left( \frac{x}{a} \right) - \sin^{-1} \left( \frac{y}{b} \right) = \theta \]
Using the identity \( \sin^{-1}(x) = \frac{\pi}{2} - \cos^{-1}(x) \), we can rewrite the equation:
\[ \cos^{-1} \left( \frac{x}{a} \right) + \cos^{-1} \left( \frac{y}{b} \right) = \frac{\pi}{2} + \theta \]

### Step 2: Apply the cosine addition formula
Using the cosine addition formula:
\[ \cos^{-1}(u) + \cos^{-1}(v) = \cos^{-1}(uv - \sqrt{(1-u^2)(1-v^2)}) \]
we set \( u = \frac{x}{a} \) and \( v = \frac{y}{b} \):
\[ \cos^{-1} \left( \frac{x}{a} \cdot \frac{y}{b} - \sqrt{\left(1 - \left(\frac{x}{a}\right)^2\right)\left(1 - \left(\frac{y}{b}\right)^2\right)} \right) = \frac{\pi}{2} + \theta \]

### Step 3: Take cosine of both sides
Taking cosine of both sides:
\[ \frac{x}{a} \cdot \frac{y}{b} - \sqrt{\left(1 - \left(\frac{x}{a}\right)^2\right)\left(1 - \left(\frac{y}{b}\right)^2\right)} = -\sin(\theta) \]

### Step 4: Rearranging the equation
Rearranging gives:
\[ \frac{x}{a} \cdot \frac{y}{b} + \sin(\theta) = \sqrt{\left(1 - \left(\frac{x}{a}\right)^2\right)\left(1 - \left(\frac{y}{b}\right)^2\right)} \]

### Step 5: Square both sides
Squaring both sides:
\[ \left(\frac{x}{a} \cdot \frac{y}{b} + \sin(\theta)\right)^2 = \left(1 - \left(\frac{x}{a}\right)^2\right)\left(1 - \left(\frac{y}{b}\right)^2\right) \]

### Step 6: Expand and simplify
Expanding both sides leads to:
\[ \frac{x^2 y^2}{a^2 b^2} + 2\frac{xy}{ab}\sin(\theta) + \sin^2(\theta) = 1 - \frac{x^2}{a^2} - \frac{y^2}{b^2} + \frac{x^2 y^2}{a^2 b^2} \]
This simplifies to:
\[ 2\frac{xy}{ab}\sin(\theta) + \sin^2(\theta) = 1 - \frac{x^2}{a^2} - \frac{y^2}{b^2} \]

### Step 7: Rearranging to find the maximum value
We can rearrange to express \( b^2 x^2 + a^2 y^2 + 2ab xy \sin(\theta) \):
\[ b^2 x^2 + a^2 y^2 + 2ab xy \sin(\theta) = a^2 b^2 \cos^2(\theta) \]

### Step 8: Finding the maximum value
The maximum value occurs when \( \cos^2(\theta) \) is maximized, which is 1 when \( \theta = 0 \):
\[ \text{Maximum Value} = a^2 b^2 \]

Thus, the maximum value of \( b^2 x^2 + a^2 y^2 + 2ab xy \sin(\theta) \) is \( a^2 b^2 \).

### Final Answer
The maximum value is \( a^2 b^2 \).

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