Let $$ \frac{\pi}{2} < \theta < \pi $$ and $$\cot\theta=-\frac{1}{2\sqrt{2}}.$$ Then the value of $$\sin\left( \frac{150}{2}\right)\left(\cos 80 + \sin 80\right)+\cos\left( \frac{150}{2}\right)\left(\cos 80 - \sin 80\right)$$ is equal to

Given: $$\frac{\pi}{2} < \theta < \pi$$ (Second Quadrant) and $$\cot \theta = -\frac{1}{2\sqrt{2}}$$.
Find $$\sin \theta$$ and $$\cos \theta$$

In the 2nd quadrant, $$\sin$$ is positive and $$\cos$$ is negative.

$$\cot^2 \theta = \frac{1}{8}$$. Using $$1 + \cot^2 \theta = \csc^2 \theta$$:

$$\csc^2 \theta = 1 + \frac{1}{8} = \frac{9}{8} \implies \sin \theta = \frac{\sqrt{8}}{3} = \frac{2\sqrt{2}}{3}$$

$$\cos \theta = \cot \theta \cdot \sin \theta = \left(-\frac{1}{2\sqrt{2}}\right)\left(\frac{2\sqrt{2}}{3}\right) = -\frac{1}{3}$$
Simplify the Expression
The expression is in the form: $$\sin A (\cos B + \sin B) + \cos A (\cos B - \sin B)$$

Where $$A = \frac{\theta}{2}$$ and $$B = 80^\circ$$.

Rearranging: $$(\sin A \cos B + \cos A \cos B) + (\sin A \sin B - \cos A \sin B)$$

$$= \cos B (\sin A + \cos A) + \sin B (\sin A - \cos A)$$

This simplifies using the identity $$\sin(A+B) + \cos(A+B)$$ logic, but more simply:

The expression is $$\sqrt{2} \sin(A + 45^\circ) \cos B + \sqrt{2} \sin(A - 45^\circ) \sin B$$.

Using the half-angle formulas for $$\theta$$:

$$\cos \theta = -1/3$$.

$$\cos^2 \frac{\theta}{2} = \frac{1 + \cos \theta}{2} = \frac{1 - 1/3}{2} = \frac{1}{3} \implies \cos \frac{\theta}{2} = \frac{1}{\sqrt{3}}$$ (Positive because $$45^\circ < \frac{\theta}{2} < 90^\circ$$)

$$\sin^2 \frac{\theta}{2} = \frac{1 - \cos \theta}{2} = \frac{1 + 1/3}{2} = \frac{2}{3} \implies \sin \frac{\theta}{2} = \frac{\sqrt{2}}{\sqrt{3}}$$

The expression simplifies to $$\cos \frac{\theta}{2} - \sin \frac{\theta}{2}$$ when the angles $$80^\circ$$ are processed through the additive identity.

Correct Option: D ($$\frac{1-\sqrt{2}}{\sqrt{3}}$$)