Without using tables, evaluate the following:
`3 cos 68^(@) . "cosec" 22^(@) - (1)/(2) tan 43^(@) .tan 47^(@) .tan 12^(@).tan 60^(@).tan 78^(@)`.

Answer

To evaluate the expression \( 3 \cos 68^\circ \cdot \csc 22^\circ - \frac{1}{2} \tan 43^\circ \cdot \tan 47^\circ \cdot \tan 12^\circ \cdot \tan 60^\circ \cdot \tan 78^\circ \), we can follow these steps:

### Step 1: Simplify \( 3 \cos 68^\circ \cdot \csc 22^\circ \)

We know that:
\[ \csc 22^\circ = \sec(90^\circ - 22^\circ) = \sec 68^\circ \]
Thus, we can rewrite the expression:
\[ 3 \cos 68^\circ \cdot \csc 22^\circ = 3 \cos 68^\circ \cdot \sec 68^\circ \]
Since \( \sec \theta = \frac{1}{\cos \theta} \), we have:
\[ 3 \cos 68^\circ \cdot \sec 68^\circ = 3 \cos 68^\circ \cdot \frac{1}{\cos 68^\circ} = 3 \]

### Step 2: Simplify \( \frac{1}{2} \tan 43^\circ \cdot \tan 47^\circ \cdot \tan 12^\circ \cdot \tan 60^\circ \cdot \tan 78^\circ \)

Using the identity \( \tan(90^\circ - \theta) = \cot \theta \), we find:
\[ \tan 47^\circ = \cot 43^\circ \]
Thus:
\[ \tan 43^\circ \cdot \tan 47^\circ = \tan 43^\circ \cdot \cot 43^\circ = 1 \]
Next, we simplify \( \tan 60^\circ \) and \( \tan 78^\circ \):
\[ \tan 78^\circ = \cot 12^\circ \]
So:
\[ \tan 12^\circ \cdot \tan 78^\circ = \tan 12^\circ \cdot \cot 12^\circ = 1 \]
Now, we can substitute these values back into the expression:
\[ \frac{1}{2} \cdot 1 \cdot 1 = \frac{1}{2} \]

### Step 3: Combine the results

Now we substitute back into the original expression:
\[ 3 - \frac{1}{2} \]
To combine these, we convert 3 into a fraction:
\[ 3 = \frac{6}{2} \]
Thus:
\[ \frac{6}{2} - \frac{1}{2} = \frac{6 - 1}{2} = \frac{5}{2} \]

### Final Answer

The final answer is:
\[ \frac{5}{2} \]