1. Stating the problem: Prove the identity $$\csc 2A = \frac{1}{2}(\sec A \csc A)$$. 2. Recall the double angle identity for cosecant: $$\csc 2A = \frac{1}{\sin 2A}$$. 3. Recall that $$\sin 2A = 2 \sin A \cos A$$, so: $$\csc 2A = \frac{1}{2 \sin A \cos A}$$. 4. Now, write $$\sec A$$ and $$\csc A$$ in terms of sine and cosine: $$\sec A = \frac{1}{\cos A}, \quad \csc A = \frac{1}{\sin A}$$. 5. Compute the right hand side (RHS): $$\frac{1}{2}(\sec A \csc A) = \frac{1}{2} \times \frac{1}{\cos A} \times \frac{1}{\sin A} = \frac{1}{2 \sin A \cos A}$$. 6. Therefore, both sides simplify to: $$\csc 2A = \frac{1}{2 \sin A \cos A}$$ and $$\frac{1}{2}(\sec A \csc A) = \frac{1}{2 \sin A \cos A}$$. 7. Hence, the identity is proven: $$\csc 2A = \frac{1}{2}(\sec A \csc A)$$.