If x and y are the solutions of the equation `12sinx+5cos x=2y^(2)-8y+21`, then the value of `12cot((xy)/(2))` is (Given, `|x| lt pi`)

Answer

To solve the equation \(12\sin x + 5\cos x = 2y^2 - 8y + 21\) and find the value of \(12\cot\left(\frac{xy}{2}\right)\), we can follow these steps:

### Step 1: Determine the Range of the Left-Hand Side (LHS)
The LHS of the equation is \(12\sin x + 5\cos x\). We can find its maximum and minimum values using the formula for the maximum value of \(a \sin x + b \cos x\), which is given by \(\sqrt{a^2 + b^2}\).

Here, \(a = 12\) and \(b = 5\):
\[ \text{Maximum value} = \sqrt{12^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = 13 \]
Thus, the range of \(12\sin x + 5\cos x\) is \([-13, 13]\).

### Step 2: Analyze the Right-Hand Side (RHS)
The RHS of the equation is \(2y^2 - 8y + 21\). We can rewrite this quadratic expression in vertex form to find its minimum value.

Completing the square:
\[ 2y^2 - 8y + 21 = 2(y^2 - 4y) + 21 = 2(y^2 - 4y + 4 - 4) + 21 = 2((y - 2)^2 - 4) + 21 \]
\[ = 2(y - 2)^2 - 8 + 21 = 2(y - 2)^2 + 13 \]
The minimum value occurs when \((y - 2)^2 = 0\), i.e., \(y = 2\). Therefore, the minimum value of the RHS is \(13\).

### Step 3: Set the LHS Equal to the RHS
For the equation \(12\sin x + 5\cos x = 2y^2 - 8y + 21\) to have solutions, the maximum value of the LHS must equal the minimum value of the RHS:
\[ 12\sin x + 5\cos x = 13 \quad \text{and} \quad 2y^2 - 8y + 21 = 13 \]
Setting the RHS equal to 13:
\[ 2y^2 - 8y + 21 = 13 \implies 2y^2 - 8y + 8 = 0 \implies y^2 - 4y + 4 = 0 \implies (y - 2)^2 = 0 \implies y = 2 \]

### Step 4: Solve for \(x\)
Now substituting \(y = 2\) back into the LHS:
\[ 12\sin x + 5\cos x = 13 \]
Dividing the entire equation by 13:
\[ \frac{12}{13}\sin x + \frac{5}{13}\cos x = 1 \]
This can be rewritten in the form of \(R\sin(x + \alpha)\):
\[ R = \sqrt{\left(\frac{12}{13}\right)^2 + \left(\frac{5}{13}\right)^2} = \sqrt{\frac{144 + 25}{169}} = \sqrt{1} = 1 \]
Thus, we can express it as:
\[ \sin\left(x + \alpha\right) = 1 \]
where \(\tan \alpha = \frac{5/13}{12/13} = \frac{5}{12}\).

### Step 5: Find \(x\)
The equation \(\sin\left(x + \alpha\right) = 1\) implies:
\[ x + \alpha = \frac{\pi}{2} \implies x = \frac{\pi}{2} - \alpha \]
Calculating \(\alpha\):
\[ \alpha = \tan^{-1}\left(\frac{5}{12}\right) \]
Thus,
\[ x = \frac{\pi}{2} - \tan^{-1}\left(\frac{5}{12}\right) \]

### Step 6: Calculate \(xy\)
Now, we need to find \(xy\):
\[ xy = 2\left(\frac{\pi}{2} - \tan^{-1}\left(\frac{5}{12}\right)\right) = \pi - 2\tan^{-1}\left(\frac{5}{12}\right) \]

### Step 7: Find \(12\cot\left(\frac{xy}{2}\right)\)
Now, we need to find:
\[ \cot\left(\frac{xy}{2}\right) = \cot\left(\frac{\pi}{2} - \tan^{-1}\left(\frac{5}{12}\right)\right) = \tan\left(\tan^{-1}\left(\frac{5}{12}\right)\right) = \frac{5}{12} \]
Thus,
\[ 12\cot\left(\frac{xy}{2}\right) = 12 \cdot \frac{5}{12} = 5 \]

### Final Answer
The value of \(12\cot\left(\frac{xy}{2}\right)\) is \(\boxed{5}\).