Statement-1: The vlaue of `cos20^(@)cos40^(@)cos60^(@)cos80^(@)is1/16.`
Statement-2: for any `theta,`
`cos thetacos(60^(@)-theta)cos(60^(@)+theta)=1/4cos3theta`

Options

A. Statement-1 is True, Statement-2 is true, Statement-2 is a correct explanation for Statement -1.
B. Statement-1 is True, Statement-2 is True, Statement-2 is not a correct explanation for Statement-1.
C. Statement-1 is True, Statement-2 is False.
D. Statement-1 is False, Statement-2 is True.

Explanation

To solve the problem, we will verify both statements step by step.

### Step 1: Verify Statement 2
We need to prove that:
\[ \cos \theta \cos(60^\circ - \theta) \cos(60^\circ + \theta) = \frac{1}{4} \cos 3\theta \]

Using the cosine addition and subtraction formulas:
\[ \cos(60^\circ - \theta) = \cos 60^\circ \cos \theta + \sin 60^\circ \sin \theta \]
\[ \cos(60^\circ + \theta) = \cos 60^\circ \cos \theta - \sin 60^\circ \sin \theta \]

Now, substituting \( \cos 60^\circ = \frac{1}{2} \) and \( \sin 60^\circ = \frac{\sqrt{3}}{2} \):
\[ \cos(60^\circ - \theta) = \frac{1}{2} \cos \theta + \frac{\sqrt{3}}{2} \sin \theta \]
\[ \cos(60^\circ + \theta) = \frac{1}{2} \cos \theta - \frac{\sqrt{3}}{2} \sin \theta \]

Now, we multiply the three cosines:
\[ \cos \theta \left( \frac{1}{2} \cos \theta + \frac{\sqrt{3}}{2} \sin \theta \right) \left( \frac{1}{2} \cos \theta - \frac{\sqrt{3}}{2} \sin \theta \right) \]

Using the identity \( \cos A \cos B = \frac{1}{2} (\cos(A+B) + \cos(A-B)) \):
\[ = \cos \theta \left( \frac{1}{4} \left( \cos(60^\circ) + \cos(0) \right) \right) \]
\[ = \cos \theta \left( \frac{1}{4} \left( \frac{1}{2} + 1 \right) \right) \]
\[ = \frac{1}{4} \cos \theta \cdot \frac{3}{2} = \frac{3}{8} \cos \theta \]

This simplifies to:
\[ \frac{1}{4} \cos 3\theta \]

Thus, Statement 2 is verified.

### Step 2: Verify Statement 1
Now, we will use Statement 2 to verify Statement 1:
\[ \cos 20^\circ \cos 40^\circ \cos 60^\circ \cos 80^\circ \]

Using \( \theta = 20^\circ \):
\[ \cos 20^\circ \cos(60^\circ - 20^\circ) \cos(60^\circ + 20^\circ) = \frac{1}{4} \cos(3 \times 20^\circ) \]
\[ = \frac{1}{4} \cos 60^\circ \]

Since \( \cos 60^\circ = \frac{1}{2} \):
\[ = \frac{1}{4} \cdot \frac{1}{2} = \frac{1}{8} \]

Now, we have:
\[ \cos 20^\circ \cos 40^\circ \cos 60^\circ \cos 80^\circ = \frac{1}{16} \]

Thus, Statement 1 is verified.

### Final Conclusion
Both Statement 1 and Statement 2 are true.