1. **State the problem:** Calculate the value of $$\sin\left(-\frac{\pi}{12}\right) \cdot \csc\left(\frac{25\pi}{12}\right)$$. 2. **Recall definitions and properties:** - $$\csc x = \frac{1}{\sin x}$$. - The sine function is periodic with period $$2\pi$$. - $$\sin(-x) = -\sin x$$. 3. **Simplify the expression:** $$\sin\left(-\frac{\pi}{12}\right) \cdot \csc\left(\frac{25\pi}{12}\right) = \sin\left(-\frac{\pi}{12}\right) \cdot \frac{1}{\sin\left(\frac{25\pi}{12}\right)}$$ 4. **Use the odd function property of sine:** $$\sin\left(-\frac{\pi}{12}\right) = -\sin\left(\frac{\pi}{12}\right)$$ 5. **Reduce the angle $$\frac{25\pi}{12}$$ modulo $$2\pi$$:** Since $$2\pi = \frac{24\pi}{12}$$, $$\frac{25\pi}{12} = 2\pi + \frac{\pi}{12}$$, so $$\sin\left(\frac{25\pi}{12}\right) = \sin\left(2\pi + \frac{\pi}{12}\right) = \sin\left(\frac{\pi}{12}\right)$$ 6. **Substitute back:** $$-\sin\left(\frac{\pi}{12}\right) \cdot \frac{1}{\sin\left(\frac{\pi}{12}\right)}$$ 7. **Cancel common factors:** $$-\cancel{\sin\left(\frac{\pi}{12}\right)} \cdot \frac{1}{\cancel{\sin\left(\frac{\pi}{12}\right)}} = -1$$ **Final answer:** $$\boxed{-1}$$