We are tasked with finding the value of \( \cos 1^\circ \cos 2^\circ \cdots \cos 180^\circ \).
Step 1: Analyze the product.
The product involves all cosine values from \( \cos 1^\circ \) to \( \cos 180^\circ \). A key observation is that \( \cos 90^\circ = 0 \).
Step 2: Impact of \( \cos 90^\circ = 0 \).
Since one of the terms in the product is \( \cos 90^\circ = 0 \), the entire product becomes zero. This is because multiplying any number by zero results in zero.
Final Answer: The value of \( \cos 1^\circ \cos 2^\circ \cdots \cos 180^\circ \) is \( \mathbf{0} \).
The value of \(\dfrac{\sqrt{3}\cosec 20^\circ - \sec 20^\circ}{\cos 20^\circ \cos 40^\circ \cos 60^\circ \cos 80^\circ}\) is equal to
If $\cot x=\dfrac{5}{12}$ for some $x\in(\pi,\tfrac{3\pi}{2})$, then \[ \sin 7x\left(\cos \frac{13x}{2}+\sin \frac{13x}{2}\right) +\cos 7x\left(\cos \frac{13x}{2}-\sin \frac{13x}{2}\right) \] is equal to
If \[ \frac{\cos^2 48^\circ - \sin^2 12^\circ}{\sin^2 24^\circ - \sin^2 6^\circ} = \frac{\alpha + \beta\sqrt{5}}{2}, \] where \( \alpha, \beta \in \mathbb{N} \), then the value of \( \alpha + \beta \) is ___________.