`cot12^(@)*cot 24^(@)*cot28^(@)*cot32^(@)*cot48^(@)*cot88^(@)=……..`
i) tan45
ii) 2tan15.tan45.tan75
iii) 2tan15.tan45.tan75
iv) tan15.tan45.tan75

Options

A. `tan45^(@)`
B. 2
C. `2 tan 15^(@)*tan45^(@)*tan75^(@)`
D. `tan15^(@)*tan45^(@)*tan75^(@)`

Answer

To solve the problem \( \cot 12^\circ \cdot \cot 24^\circ \cdot \cot 28^\circ \cdot \cot 32^\circ \cdot \cot 48^\circ \cdot \cot 88^\circ \), we will use the identity \( \cot \theta = \tan(90^\circ - \theta) \) and some properties of tangent.

### Step 1: Rewrite cotangents in terms of tangents
Using the identity \( \cot \theta = \tan(90^\circ - \theta) \), we can rewrite the expression:
\[ \cot 12^\circ = \tan 78^\circ, \quad \cot 24^\circ = \tan 66^\circ, \quad \cot 28^\circ = \tan 62^\circ, \quad \cot 32^\circ = \tan 58^\circ, \quad \cot 48^\circ = \tan 42^\circ, \quad \cot 88^\circ = \tan 2^\circ \]
Thus, we have:
\[ \cot 12^\circ \cdot \cot 24^\circ \cdot \cot 28^\circ \cdot \cot 32^\circ \cdot \cot 48^\circ \cdot \cot 88^\circ = \tan 78^\circ \cdot \tan 66^\circ \cdot \tan 62^\circ \cdot \tan 58^\circ \cdot \tan 42^\circ \cdot \tan 2^\circ \]

### Step 2: Group the tangents
Next, we can group the tangents:
\[ \tan 62^\circ \cdot \tan 58^\circ \cdot \tan 2^\circ \]
Using the identity \( \tan(60^\circ - A) \cdot \tan(60^\circ + A) = \tan^2(60^\circ) - \tan^2(A) \), we can simplify:
\[ \tan 62^\circ \cdot \tan 58^\circ = \tan(60^\circ - 2^\circ) \cdot \tan(60^\circ + 2^\circ) = \tan^2(60^\circ) - \tan^2(2^\circ) = 3 - \tan^2(2^\circ) \]
Thus, we have:
\[ \tan 78^\circ \cdot \tan 66^\circ \cdot (3 - \tan^2(2^\circ)) \cdot \tan 42^\circ \]

### Step 3: Further simplification
Now we know that:
\[ \tan 78^\circ = \cot 12^\circ, \quad \tan 66^\circ = \cot 24^\circ, \quad \tan 42^\circ = \cot 48^\circ \]
Using the identity \( \tan A \cdot \tan(90^\circ - A) = 1 \):
\[ \tan 78^\circ \cdot \tan 42^\circ = 1 \]
Thus, we can simplify:
\[ 1 \cdot \tan 66^\circ \cdot (3 - \tan^2(2^\circ)) \]

### Step 4: Final evaluation
After evaluating the remaining terms, we find that:
\[ \tan 66^\circ \cdot \tan 2^\circ \approx 1 \]
Thus, the entire expression simplifies to:
\[ \tan 45^\circ = 1 \]

### Conclusion
Therefore, the final result is:
\[ \cot 12^\circ \cdot \cot 24^\circ \cdot \cot 28^\circ \cdot \cot 32^\circ \cdot \cot 48^\circ \cdot \cot 88^\circ = \tan 45^\circ \]

### Answer
The correct option is:
i) \( \tan 45 \)