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To differentiate the expression $$\frac{5 + 3\cos(\theta)}{6\sin(\theta)}$$ with respect to $$\theta$$, we will use the quotient rule. The quotient rule for differentiation states that if we have a function $$f(\theta) = \frac{u(\theta)}{v(\theta)}$$, then its derivative is given by:

$f'(\theta) = \frac{u'(\theta)v(\theta) - u(\theta)v'(\theta)}{[v(\theta)]^2}$

### Step-by-Step Differentiation

1. **Identify $$u(\theta)$$ and $$v(\theta)$$:**
- $$u(\theta) = 5 + 3\cos(\theta)$$
- $$v(\theta) = 6\sin(\theta)$$

2. **Differentiate $$u(\theta)$$ and $$v(\theta)$$:**
- $$u'(\theta) = \frac{d}{d\theta} [5 + 3\cos(\theta)]$$
$u'(\theta) = 0 + 3(-\sin(\theta)) = -3\sin(\theta)$

- $$v'(\theta) = \frac{d}{d\theta} [6\sin(\theta)]$$
$v'(\theta) = 6\cos(\theta)$

3. **Apply the quotient rule:**
$f'(\theta) = \frac{u'(\theta)v(\theta) - u(\theta)v'(\theta)}{[v(\theta)]^2}$
Substitute $$u(\theta)$$, $$v(\theta)$$, $$u'(\theta)$$, and $$v'(\theta)$$:
$f'(\theta) = \frac{(-3\sin(\theta))(6\sin(\theta)) - (5 + 3\cos(\theta))(6\cos(\theta))}{[6\sin(\theta)]^2}$

4. **Simplify the numerator:**
$\text{Numerator} = (-3\sin(\theta))(6\sin(\theta)) - (5 + 3\cos(\theta))(6\cos(\theta))$
$= -18\sin^2(\theta) - (30\cos(\theta) + 18\cos^2(\theta))$
$= -18\sin^2(\theta) - 30\cos(\theta) - 18\cos^2(\theta)$

5. **Simplify the denominator:**
$\text{Denominator} = [6\sin(\theta)]^2 = 36\sin^2(\theta)$

6. **Combine the numerator and denominator:**
$f'(\theta) = \frac{-18\sin^2(\theta) - 30\cos(\theta) - 18\cos^2(\theta)}{36\sin^2(\theta)}$

7. **Factor and simplify if possible:**
$f'(\theta) = \frac{-18(\sin^2(\theta) + \cos^2(\theta)) - 30\cos(\theta)}{36\sin^2(\theta)}$
Using the Pythagorean identity $$\sin^2(\theta) + \cos^2(\theta) = 1$$:
$f'(\theta) = \frac{-18(1) - 30\cos(\theta)}{36\sin^2(\theta)}$
$= \frac{-18 - 30\cos(\theta)}{36\sin^2(\theta)}$
Simplify the fraction:
$= \frac{-18 - 30\cos(\theta)}{36\sin^2(\theta)} = \frac{-18(1 + \frac{5\cos(\theta)}{3})}{36\sin^2(\theta)}$
$= \frac{-18(3 + 5\cos(\theta))}{108\sin^2(\theta)} = \frac{-3(3 + 5\cos(\theta))}{18\sin^2(\theta)}$
$= \frac{- (3 + 5\cos(\theta))}{6\sin^2(\theta)}$

So, the derivative of $$\frac{5 + 3\cos(\theta)}{6\sin(\theta)}$$ with respect to $$\theta$$ is:

$f'(\theta) = \frac{-(3 + 5\cos(\theta))}{6\sin^2(\theta)}$