If $$\operatorname{cosec} \theta = \frac{p+q}{p-q}$$ ($$p \neq q, p \neq 0$$), then $$\left|\cot\left(\frac{\pi}{4} + \frac{\theta}{2}\right)\right|$$ is equals to:

We are given that $$\cosec \theta = \frac{p+q}{p-q}$$ with $$p \neq q$$ and $$p \neq 0$$. We need to find $$\left|\cot\left(\frac{\pi}{4} + \frac{\theta}{2}\right)\right|$$.

First, recall that $$\cosec \theta = \frac{1}{\sin \theta}$$, so $$\sin \theta = \frac{p-q}{p+q}$$.

Now, consider the expression $$\cot\left(\frac{\pi}{4} + \frac{\theta}{2}\right)$$. Using the angle addition formula for cotangent, $$\cot(A + B) = \frac{\cot A \cot B - 1}{\cot B + \cot A}$$, set $$A = \frac{\pi}{4}$$ and $$B = \frac{\theta}{2}$$. Since $$\cot \frac{\pi}{4} = 1$$, we have:

$$ \cot\left(\frac{\pi}{4} + \frac{\theta}{2}\right) = \frac{\cot \frac{\pi}{4} \cot \frac{\theta}{2} - 1}{\cot \frac{\theta}{2} + \cot \frac{\pi}{4}} = \frac{(1) \cdot \cot \frac{\theta}{2} - 1}{\cot \frac{\theta}{2} + 1} = \frac{\cot \frac{\theta}{2} - 1}{\cot \frac{\theta}{2} + 1}. $$

Let $$u = \cot \frac{\theta}{2}$$. So the expression becomes $$\frac{u - 1}{u + 1}$$.

Next, express $$\sin \theta$$ in terms of $$u$$. We know that $$\sin \theta = 2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}$$. Since $$u = \cot \frac{\theta}{2} = \frac{\cos \frac{\theta}{2}}{\sin \frac{\theta}{2}}$$, set $$t = \tan \frac{\theta}{2}$$, so $$u = \frac{1}{t}$$. Then $$\sin \theta = \frac{2t}{1+t^2}$$. Substituting $$t = \frac{1}{u}$$:

$$ \sin \theta = \frac{2 \cdot \frac{1}{u}}{1 + \left(\frac{1}{u}\right)^2} = \frac{\frac{2}{u}}{\frac{u^2 + 1}{u^2}} = \frac{2}{u} \cdot \frac{u^2}{u^2 + 1} = \frac{2u}{u^2 + 1}. $$

Given $$\sin \theta = \frac{p-q}{p+q}$$, we equate:

$$ \frac{2u}{u^2 + 1} = \frac{p-q}{p+q}. $$

Cross-multiplying:

$$ 2u (p + q) = (p - q) (u^2 + 1). $$

Expanding both sides:

$$ 2u(p + q) = (p - q)u^2 + (p - q). $$

Rearranging all terms to one side:

$$ (p - q)u^2 - 2(p + q)u + (p - q) = 0. $$

This is a quadratic equation in $$u$$. Solving using the quadratic formula:

$$ u = \frac{2(p + q) \pm \sqrt{[-2(p + q)]^2 - 4(p - q)(p - q)}}{2(p - q)}. $$

Simplifying the discriminant:

$$ [-2(p + q)]^2 = 4(p + q)^2, $$

$$ 4(p - q)(p - q) = 4(p - q)^2, $$

so

$$ \text{discriminant} = 4(p + q)^2 - 4(p - q)^2 = 4[(p + q)^2 - (p - q)^2]. $$

Now, $$(p + q)^2 - (p - q)^2 = (p^2 + 2pq + q^2) - (p^2 - 2pq + q^2) = 4pq$$, so:

$$ \text{discriminant} = 4 \cdot 4pq = 16pq. $$

Thus,

$$ u = \frac{2(p + q) \pm \sqrt{16pq}}{2(p - q)} = \frac{2(p + q) \pm 4\sqrt{pq}}{2(p - q)} = \frac{(p + q) \pm 2\sqrt{pq}}{p - q}. $$

Note that $$(p + q) \pm 2\sqrt{pq} = (\sqrt{p} \pm \sqrt{q})^2$$ and $$p - q = (\sqrt{p})^2 - (\sqrt{q})^2 = (\sqrt{p} - \sqrt{q})(\sqrt{p} + \sqrt{q})$$. So:

$$ u = \frac{(\sqrt{p} \pm \sqrt{q})^2}{(\sqrt{p} - \sqrt{q})(\sqrt{p} + \sqrt{q})}. $$

This gives two cases:

Case 1: Using the positive sign:

$$ u_1 = \frac{(\sqrt{p} + \sqrt{q})^2}{(\sqrt{p} - \sqrt{q})(\sqrt{p} + \sqrt{q})} = \frac{\sqrt{p} + \sqrt{q}}{\sqrt{p} - \sqrt{q}}. $$

Case 2: Using the negative sign:

$$ u_2 = \frac{(\sqrt{p} - \sqrt{q})^2}{(\sqrt{p} - \sqrt{q})(\sqrt{p} + \sqrt{q})} = \frac{\sqrt{p} - \sqrt{q}}{\sqrt{p} + \sqrt{q}}. $$

Now, substitute back into $$\frac{u - 1}{u + 1}$$.

For $$u_1 = \frac{\sqrt{p} + \sqrt{q}}{\sqrt{p} - \sqrt{q}}$$:

$$ \frac{u_1 - 1}{u_1 + 1} = \frac{\frac{\sqrt{p} + \sqrt{q}}{\sqrt{p} - \sqrt{q}} - 1}{\frac{\sqrt{p} + \sqrt{q}}{\sqrt{p} - \sqrt{q}} + 1}. $$

Simplify numerator:

$$ \frac{\sqrt{p} + \sqrt{q}}{\sqrt{p} - \sqrt{q}} - 1 = \frac{\sqrt{p} + \sqrt{q} - (\sqrt{p} - \sqrt{q})}{\sqrt{p} - \sqrt{q}} = \frac{2\sqrt{q}}{\sqrt{p} - \sqrt{q}}. $$

Simplify denominator:

$$ \frac{\sqrt{p} + \sqrt{q}}{\sqrt{p} - \sqrt{q}} + 1 = \frac{\sqrt{p} + \sqrt{q} + \sqrt{p} - \sqrt{q}}{\sqrt{p} - \sqrt{q}} = \frac{2\sqrt{p}}{\sqrt{p} - \sqrt{q}}. $$

So:

$$ \frac{u_1 - 1}{u_1 + 1} = \frac{\frac{2\sqrt{q}}{\sqrt{p} - \sqrt{q}}}{\frac{2\sqrt{p}}{\sqrt{p} - \sqrt{q}}} = \frac{2\sqrt{q}}{2\sqrt{p}} = \sqrt{\frac{q}{p}}. $$

For $$u_2 = \frac{\sqrt{p} - \sqrt{q}}{\sqrt{p} + \sqrt{q}}$$:

$$ \frac{u_2 - 1}{u_2 + 1} = \frac{\frac{\sqrt{p} - \sqrt{q}}{\sqrt{p} + \sqrt{q}} - 1}{\frac{\sqrt{p} - \sqrt{q}}{\sqrt{p} + \sqrt{q}} + 1}. $$

Simplify numerator:

$$ \frac{\sqrt{p} - \sqrt{q}}{\sqrt{p} + \sqrt{q}} - 1 = \frac{\sqrt{p} - \sqrt{q} - (\sqrt{p} + \sqrt{q})}{\sqrt{p} + \sqrt{q}} = \frac{-2\sqrt{q}}{\sqrt{p} + \sqrt{q}}. $$

Simplify denominator:

$$ \frac{\sqrt{p} - \sqrt{q}}{\sqrt{p} + \sqrt{q}} + 1 = \frac{\sqrt{p} - \sqrt{q} + \sqrt{p} + \sqrt{q}}{\sqrt{p} + \sqrt{q}} = \frac{2\sqrt{p}}{\sqrt{p} + \sqrt{q}}. $$

So:

$$ \frac{u_2 - 1}{u_2 + 1} = \frac{\frac{-2\sqrt{q}}{\sqrt{p} + \sqrt{q}}}{\frac{2\sqrt{p}}{\sqrt{p} + \sqrt{q}}} = \frac{-2\sqrt{q}}{2\sqrt{p}} = -\sqrt{\frac{q}{p}}. $$

Thus, $$\cot\left(\frac{\pi}{4} + \frac{\theta}{2}\right)$$ is either $$\sqrt{\frac{q}{p}}$$ or $$-\sqrt{\frac{q}{p}}$$. Taking the absolute value:

$$ \left|\cot\left(\frac{\pi}{4} + \frac{\theta}{2}\right)\right| = \left| \pm \sqrt{\frac{q}{p}} \right| = \sqrt{\frac{q}{p}}. $$

Comparing with the options:

A. $$pq$$

B. $$\sqrt{\frac{p}{q}}$$

C. $$\sqrt{\frac{q}{p}}$$

D. $$\sqrt{pq}$$

Option C matches $$\sqrt{\frac{q}{p}}$$. Hence, the correct answer is Option C.