The value of `(cot84^(@)cot48^(@))/(cot66^(@)cot78^(@))` is equal to

Options

A. 1
B. 0
C. `sqrt(3)`
D. none of these

Explanation

To solve the expression \(\frac{\cot 84^\circ \cot 48^\circ}{\cot 66^\circ \cot 78^\circ}\), we can follow these steps:

### Step 1: Rewrite cotangent in terms of sine and cosine
We know that \(\cot x = \frac{\cos x}{\sin x}\). Therefore, we can rewrite the expression as:
\[ \frac{\cot 84^\circ \cot 48^\circ}{\cot 66^\circ \cot 78^\circ} = \frac{\frac{\cos 84^\circ}{\sin 84^\circ} \cdot \frac{\cos 48^\circ}{\sin 48^\circ}}{\frac{\cos 66^\circ}{\sin 66^\circ} \cdot \frac{\cos 78^\circ}{\sin 78^\circ}} \]

### Step 2: Combine the fractions
This can be simplified to:
\[ \frac{\cos 84^\circ \cos 48^\circ \sin 66^\circ \sin 78^\circ}{\sin 84^\circ \sin 48^\circ \cos 66^\circ \cos 78^\circ} \]

### Step 3: Use the sine and cosine double angle identities
We can use the identities \(2 \sin A \cos B = \sin(A + B) + \sin(A - B)\) and \(2 \cos A \sin B = \sin(A + B) - \sin(A - B)\) to simplify the expression further.

### Step 4: Apply the identities
We will apply the identities to the numerator and denominator:
1. For \(2 \cos 84^\circ \sin 66^\circ\):
\[ 2 \cos 84^\circ \sin 66^\circ = \sin(84 + 66) + \sin(84 - 66) = \sin 150^\circ + \sin 18^\circ \]
2. For \(2 \sin 78^\circ \cos 48^\circ\):
\[ 2 \sin 78^\circ \cos 48^\circ = \sin(78 + 48) + \sin(78 - 48) = \sin 126^\circ + \sin 30^\circ \]

### Step 5: Substitute back into the expression
Now substituting back, we have:
\[ \frac{\sin 150^\circ + \sin 18^\circ}{\sin 84^\circ \sin 48^\circ \cos 66^\circ \cos 78^\circ} \]

### Step 6: Evaluate the sine values
We know:
- \(\sin 150^\circ = \sin 30^\circ = \frac{1}{2}\)
- \(\sin 18^\circ\) is a known value, approximately \(0.309\)
- \(\sin 30^\circ = \frac{1}{2}\)

### Step 7: Substitute known values
Now substituting these values into the expression:
\[ \frac{\frac{1}{2} + \sin 18^\circ}{\frac{1}{2} + \sin 18^\circ} = 1 \]

### Final Answer
Thus, the value of \(\frac{\cot 84^\circ \cot 48^\circ}{\cot 66^\circ \cot 78^\circ}\) is equal to \(1\).

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