If `cos^(-1)((x)/(3))+cos^(-1)((y)/(2))=(theta)/(2)`, then the value of `4x^(2)-12xy cos((theta)/(2))+9y^(2)` is equal to :

Options

A. `18(1+cos theta)`
B. `18(1-cos theta)`
C. `36(1+cos theta)`
D. `36(1-cos theta)`

Answer

To solve the problem step by step, we start with the given equation:

\[ \cos^{-1}\left(\frac{x}{3}\right) + \cos^{-1}\left(\frac{y}{2}\right) = \frac{\theta}{2} \]

### Step 1: Use the formula for the sum of inverse cosines
We can use the formula:
\[ \cos^{-1}(a) + \cos^{-1}(b) = \cos^{-1}(ab - \sqrt{(1-a^2)(1-b^2)}) \]
In our case, \(a = \frac{x}{3}\) and \(b = \frac{y}{2}\). Thus, we have:
\[ \cos^{-1}\left(\frac{x}{3}\right) + \cos^{-1}\left(\frac{y}{2}\right) = \cos^{-1}\left(\frac{xy}{6} - \sqrt{\left(1 - \left(\frac{x}{3}\right)^2\right)\left(1 - \left(\frac{y}{2}\right)^2\right)}\right) \]

### Step 2: Set the equation equal to \(\frac{\theta}{2}\)
From the previous step, we equate:
\[ \frac{xy}{6} - \sqrt{\left(1 - \left(\frac{x}{3}\right)^2\right)\left(1 - \left(\frac{y}{2}\right)^2\right)} = \cos\left(\frac{\theta}{2}\right) \]

### Step 3: Isolate the square root
Rearranging gives:
\[ \sqrt{\left(1 - \left(\frac{x}{3}\right)^2\right)\left(1 - \left(\frac{y}{2}\right)^2\right)} = \frac{xy}{6} - \cos\left(\frac{\theta}{2}\right) \]

### Step 4: Square both sides
Squaring both sides results in:
\[ \left(1 - \left(\frac{x}{3}\right)^2\right)\left(1 - \left(\frac{y}{2}\right)^2\right) = \left(\frac{xy}{6} - \cos\left(\frac{\theta}{2}\right)\right)^2 \]

### Step 5: Expand both sides
Expanding the left side:
\[ 1 - \frac{x^2}{9} - \frac{y^2}{4} + \frac{x^2y^2}{36} \]
And the right side:
\[ \left(\frac{xy}{6}\right)^2 - 2\cdot\frac{xy}{6}\cdot\cos\left(\frac{\theta}{2}\right) + \cos^2\left(\frac{\theta}{2}\right) \]

### Step 6: Rearranging and simplifying
Combine all terms to one side:
\[ 1 - \frac{x^2}{9} - \frac{y^2}{4} + \frac{x^2y^2}{36} - \left(\frac{x^2y^2}{36} - \frac{xy}{3}\cos\left(\frac{\theta}{2}\right) + \cos^2\left(\frac{\theta}{2}\right)\right) = 0 \]

### Step 7: Collect like terms
This leads us to:
\[ 4x^2 + 9y^2 - 12xy\cos\left(\frac{\theta}{2}\right) + 36\cos^2\left(\frac{\theta}{2}\right) - 36 = 0 \]

### Step 8: Final expression
Rearranging gives:
\[ 4x^2 + 9y^2 - 12xy\cos\left(\frac{\theta}{2}\right) = 36(1 - \cos^2\left(\frac{\theta}{2}\right)) \]
Using the identity \(1 - \cos^2\left(\frac{\theta}{2}\right) = \sin^2\left(\frac{\theta}{2}\right)\):
\[ 4x^2 + 9y^2 - 12xy\cos\left(\frac{\theta}{2}\right) = 36\sin^2\left(\frac{\theta}{2}\right) \]

### Conclusion
Thus, the value of \(4x^2 - 12xy\cos\left(\frac{\theta}{2}\right) + 9y^2\) is equal to \(36\sin^2\left(\frac{\theta}{2}\right)\).

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