Evaluate : `sin^(2)25^(@)+sin^(2)65^(@)+sqrt(3)tan5^(@)tan15^(@)tan30^(@)tan75^(@)tan85^(@)`

Answer

To evaluate the expression \( \sin^2(25^\circ) + \sin^2(65^\circ) + \sqrt{3} \tan(5^\circ) \tan(15^\circ) \tan(30^\circ) \tan(75^\circ) \tan(85^\circ) \), we can follow these steps:

### Step 1: Simplify \( \sin^2(25^\circ) + \sin^2(65^\circ) \)

Using the identity \( \sin(90^\circ - x) = \cos(x) \), we have:
\[ \sin(65^\circ) = \cos(25^\circ) \]
Thus,
\[ \sin^2(65^\circ) = \cos^2(25^\circ) \]
Now, we can rewrite the expression:
\[ \sin^2(25^\circ) + \sin^2(65^\circ) = \sin^2(25^\circ) + \cos^2(25^\circ) \]
Using the Pythagorean identity \( \sin^2(x) + \cos^2(x) = 1 \):
\[ \sin^2(25^\circ) + \cos^2(25^\circ) = 1 \]

### Step 2: Simplify \( \sqrt{3} \tan(5^\circ) \tan(15^\circ) \tan(30^\circ) \tan(75^\circ) \tan(85^\circ) \)

Using the identity \( \tan(90^\circ - x) = \cot(x) \), we have:
\[ \tan(75^\circ) = \cot(15^\circ) \quad \text{and} \quad \tan(85^\circ) = \cot(5^\circ) \]
Thus,
\[ \tan(75^\circ) \tan(15^\circ) = 1 \quad \text{and} \quad \tan(85^\circ) \tan(5^\circ) = 1 \]
So we can rewrite:
\[ \tan(5^\circ) \tan(15^\circ) \tan(30^\circ) \tan(75^\circ) \tan(85^\circ) = \tan(5^\circ) \cdot \tan(15^\circ) \cdot \tan(30^\circ) \cdot \cot(15^\circ) \cdot \cot(5^\circ) \]
This simplifies to:
\[ \tan(30^\circ) \]
Now, we know that \( \tan(30^\circ) = \frac{1}{\sqrt{3}} \), so:
\[ \sqrt{3} \tan(5^\circ) \tan(15^\circ) \tan(30^\circ) \tan(75^\circ) \tan(85^\circ) = \sqrt{3} \cdot \frac{1}{\sqrt{3}} = 1 \]

### Step 3: Combine the results

Now we can combine the results from Step 1 and Step 2:
\[ \sin^2(25^\circ) + \sin^2(65^\circ) + \sqrt{3} \tan(5^\circ) \tan(15^\circ) \tan(30^\circ) \tan(75^\circ) \tan(85^\circ) = 1 + 1 = 2 \]

### Final Answer
Thus, the final answer is:
\[ \boxed{2} \]