Let ABC be a triangle. If `cos2A + cos2B+ cos 2C = -1` , then which one of the following is correct?

Options

A. `sinA sin B sin C=0`
B. `sinAsinBcosC=0`
C. `cosAsinBsinC=0`
D. `cosAcosBcosC=0`

Answer

To solve the problem, we need to analyze the equation given:

**Given:**
\[ \cos 2A + \cos 2B + \cos 2C = -1 \]

### Step 1: Use the angle sum property of triangles
In a triangle, the sum of angles is:
\[ A + B + C = \pi \]
From this, we can express \(C\) in terms of \(A\) and \(B\):
\[ C = \pi - A - B \]

### Step 2: Rewrite \(\cos 2C\)
Using the cosine of a sum formula:
\[ \cos 2C = \cos(2(\pi - A - B)) = \cos(2\pi - 2A - 2B) = \cos(2A + 2B) \]
Since \(\cos(2\pi - x) = \cos x\).

### Step 3: Substitute \(\cos 2C\) into the equation
Now we can substitute \(\cos 2C\) back into the original equation:
\[ \cos 2A + \cos 2B + \cos(2A + 2B) = -1 \]

### Step 4: Use the cosine addition formula
Using the cosine addition formula:
\[ \cos(2A + 2B) = \cos 2A \cos 2B - \sin 2A \sin 2B \]
Substituting this into the equation gives:
\[ \cos 2A + \cos 2B + (\cos 2A \cos 2B - \sin 2A \sin 2B) = -1 \]

### Step 5: Rearranging the equation
Rearranging the equation, we get:
\[ \cos 2A + \cos 2B + \cos 2A \cos 2B - \sin 2A \sin 2B = -1 \]
This simplifies to:
\[ \cos 2A + \cos 2B + \cos 2A \cos 2B + 1 = \sin 2A \sin 2B \]

### Step 6: Analyze the implications
From the original equation, we can derive that:
\[ \cos 2A + \cos 2B + \cos 2C = -1 \]
This implies that at least one of the angles \(A\), \(B\), or \(C\) must be such that the cosine of that angle is zero, leading to:
\[ \cos A \cos B \cos C = 0 \]
This means at least one of the angles \(A\), \(B\), or \(C\) must be \(90^\circ\) (or \(\frac{\pi}{2}\) radians).

### Conclusion
Thus, the correct conclusion from the given options is:
\[ \cos A \cos B \cos C = 0 \]