Find the value of : `((3tan41^@)/(cot49^@))^2-((sin35^@ sec55^@)/(tan10^@ tan20^@ tan60^@ tan70^@ tan80^@))^2`

Answer

To solve the expression \(\left(\frac{3 \tan 41^\circ}{\cot 49^\circ}\right)^2 - \left(\frac{\sin 35^\circ \sec 55^\circ}{\tan 10^\circ \tan 20^\circ \tan 60^\circ \tan 70^\circ \tan 80^\circ}\right)^2\), we will use trigonometric identities and properties.

### Step-by-Step Solution:

1. **Simplify \(\tan 41^\circ\) and \(\cot 49^\circ\)**:
\[ \tan 41^\circ = \cot(90^\circ - 41^\circ) = \cot 49^\circ \]
Therefore, we can substitute:
\[ \frac{3 \tan 41^\circ}{\cot 49^\circ} = \frac{3 \cot 49^\circ}{\cot 49^\circ} = 3 \]

2. **Square the result**:
\[ \left(3\right)^2 = 9 \]

3. **Simplify the second part of the expression**:
- We know that \(\sec x = \frac{1}{\cos x}\), so:
\[ \sec 55^\circ = \frac{1}{\cos 55^\circ} \]
- Also, using \(\tan x = \frac{\sin x}{\cos x}\), we can rewrite:
\[ \tan 10^\circ = \frac{\sin 10^\circ}{\cos 10^\circ}, \quad \tan 20^\circ = \frac{\sin 20^\circ}{\cos 20^\circ}, \quad \tan 60^\circ = \sqrt{3}, \quad \tan 70^\circ = \cot 20^\circ, \quad \tan 80^\circ = \cot 10^\circ \]
- Therefore:
\[ \tan 10^\circ \tan 20^\circ \tan 60^\circ \tan 70^\circ \tan 80^\circ = \tan 10^\circ \tan 20^\circ \cdot \sqrt{3} \cdot \cot 20^\circ \cdot \cot 10^\circ \]
- This simplifies to:
\[ \sqrt{3} \]

4. **Substituting into the second part**:
\[ \frac{\sin 35^\circ \sec 55^\circ}{\tan 10^\circ \tan 20^\circ \tan 60^\circ \tan 70^\circ \tan 80^\circ} = \frac{\sin 35^\circ \cdot \frac{1}{\cos 55^\circ}}{\sqrt{3}} \]

5. **Using the identity \(\sin 35^\circ = \cos 55^\circ\)**:
\[ \frac{\sin 35^\circ \cdot \frac{1}{\cos 55^\circ}}{\sqrt{3}} = \frac{\cos 55^\circ \cdot \frac{1}{\cos 55^\circ}}{\sqrt{3}} = \frac{1}{\sqrt{3}} \]

6. **Square the result**:
\[ \left(\frac{1}{\sqrt{3}}\right)^2 = \frac{1}{3} \]

7. **Combine both parts of the expression**:
\[ 9 - \frac{1}{3} = \frac{27}{3} - \frac{1}{3} = \frac{26}{3} \]

### Final Answer:
\[ \frac{26}{3} \]