Evaluate the
`((3 cos 43^(@))/(sin 47^(@)))^(2) - (cos 37^(@) "cosec" 53^(@))/(tan 5^(@) tan 25^(@) tan 45^(@) tan 65^(@) tan 85^(@))`

Answer

To evaluate the expression

\[ \left(\frac{3 \cos 43^\circ}{\sin 47^\circ}\right)^2 - \frac{\cos 37^\circ \csc 53^\circ}{\tan 5^\circ \tan 25^\circ \tan 45^\circ \tan 65^\circ \tan 85^\circ} \]

we will simplify it step by step.

### Step 1: Simplify the first term

Using the identity \(\cos(90^\circ - \theta) = \sin(\theta)\), we can rewrite \(\cos 43^\circ\) as:

\[ \cos 43^\circ = \sin(90^\circ - 43^\circ) = \sin 47^\circ \]

Thus, the first term becomes:

\[ \frac{3 \cos 43^\circ}{\sin 47^\circ} = \frac{3 \sin 47^\circ}{\sin 47^\circ} = 3 \]

Now squaring this gives:

\[ (3)^2 = 9 \]

### Step 2: Simplify the second term

For the second term, we rewrite \(\cos 37^\circ\) using the identity \(\cos(90^\circ - \theta) = \sin(\theta)\):

\[ \cos 37^\circ = \sin(90^\circ - 53^\circ) = \sin 53^\circ \]

Thus, the second term becomes:

\[ \frac{\cos 37^\circ \csc 53^\circ}{\tan 5^\circ \tan 25^\circ \tan 45^\circ \tan 65^\circ \tan 85^\circ} = \frac{\sin 53^\circ \cdot \csc 53^\circ}{\tan 5^\circ \tan 25^\circ \tan 45^\circ \tan 65^\circ \tan 85^\circ} \]

Since \(\csc 53^\circ = \frac{1}{\sin 53^\circ}\), we have:

\[ \sin 53^\circ \cdot \csc 53^\circ = 1 \]

Now we need to evaluate the denominator:

\[ \tan 5^\circ \tan 25^\circ \tan 45^\circ \tan 65^\circ \tan 85^\circ \]

Using the identity \(\tan(90^\circ - \theta) = \cot(\theta)\), we can rewrite:

\[ \tan 85^\circ = \cot 5^\circ, \quad \tan 65^\circ = \cot 25^\circ \]

Thus, we can pair them:

\[ \tan 5^\circ \tan 85^\circ = 1, \quad \tan 25^\circ \tan 65^\circ = 1 \]

And we know:

\[ \tan 45^\circ = 1 \]

So the entire product simplifies to:

\[ \tan 5^\circ \tan 25^\circ \tan 45^\circ \tan 65^\circ \tan 85^\circ = 1 \cdot 1 \cdot 1 = 1 \]

### Step 3: Combine the results

Now substituting back into the expression, we have:

\[ 9 - 1 = 8 \]

### Final Answer

Thus, the value of the expression is:

\[ \boxed{8} \]