Numerator Simplification

Let the numerator be $N = \cos^6 \theta + \sin^6 \theta + 3 \sin^2 \theta \cos^2 \theta$. We utilize the algebraic identity $a^3 + b^3 = (a+b)^3 - 3ab(a+b)$. Let $a = \cos^2 \theta$ and $b = \sin^2 \theta$. Using the fundamental trigonometric identity, $a+b = \cos^2 \theta + \sin^2 \theta = 1$. Therefore, $\cos^6 \theta + \sin^6 \theta = a^3 + b^3 = (1)^3 - 3(\cos^2 \theta)(\sin^2 \theta)(1) = 1 - 3 \sin^2 \theta \cos^2 \theta$.

Substitute this result back into the numerator expression:

$N = (1 - 3 \sin^2 \theta \cos^2 \theta) + 3 \sin^2 \theta \cos^2 \theta = 1$

Denominator Simplification

Let the denominator be $D = \csc \theta \sec \theta ((\sin \theta + \cos \theta)^2 - 1)$. First, simplify the term $\csc \theta \sec \theta$: $\csc \theta \sec \theta = \frac{1}{\sin \theta} \cdot \frac{1}{\cos \theta} = \frac{1}{\sin \theta \cos \theta}$.

Next, expand and simplify the term $(\sin \theta + \cos \theta)^2 - 1$: $(\sin \theta + \cos \theta)^2 = \sin^2 \theta + \cos^2 \theta + 2 \sin \theta \cos \theta$. Using $\sin^2 \theta + \cos^2 \theta = 1$, we get: $(\sin \theta + \cos \theta)^2 = 1 + 2 \sin \theta \cos \theta$. Therefore, $(\sin \theta + \cos \theta)^2 - 1 = (1 + 2 \sin \theta \cos \theta) - 1 = 2 \sin \theta \cos \theta$.

Now substitute these simplified terms back into the denominator expression:

$D = \left( \frac{1}{\sin \theta \cos \theta} \right) \cdot (2 \sin \theta \cos \theta) = 2$

Final Value Calculation

The value of the original trigonometric expression is obtained by dividing the simplified numerator by the simplified denominator:

Value $= \frac{N}{D} = \frac{1}{2}$