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: Rewrite \( \sin^2 65^\circ \)
We know that \( \sin(90^\circ - \theta) = \cos(\theta) \). Therefore, we can rewrite \( \sin^2 65^\circ \) as:
\[ \sin^2 65^\circ = \sin^2 (90^\circ - 25^\circ) = \cos^2 25^\circ \]

### Step 2: Combine \( \sin^2 25^\circ \) and \( \sin^2 65^\circ \)
Now, we can combine \( \sin^2 25^\circ \) and \( \sin^2 65^\circ \):
\[ \sin^2 25^\circ + \sin^2 65^\circ = \sin^2 25^\circ + \cos^2 25^\circ \]
Using the Pythagorean identity \( \sin^2 \theta + \cos^2 \theta = 1 \), we get:
\[ \sin^2 25^\circ + \cos^2 25^\circ = 1 \]

### Step 3: Rewrite the tangent terms
Next, we rewrite the tangent terms. We have:
\[ \tan 75^\circ = \tan(90^\circ - 15^\circ) = \cot 15^\circ \]
\[ \tan 85^\circ = \tan(90^\circ - 5^\circ) = \cot 5^\circ \]
Thus, we can express the product of tangents as:
\[ \tan 5^\circ \tan 15^\circ \tan 30^\circ \tan 75^\circ \tan 85^\circ = \tan 5^\circ \tan 15^\circ \tan 30^\circ \cot 15^\circ \cot 5^\circ \]

### Step 4: Simplify the tangent product
Using the identity \( \tan \theta \cot \theta = 1 \):
\[ \tan 15^\circ \cot 15^\circ = 1 \quad \text{and} \quad \tan 5^\circ \cot 5^\circ = 1 \]
Thus, we have:
\[ \tan 5^\circ \tan 15^\circ \tan 30^\circ \tan 75^\circ \tan 85^\circ = \tan 30^\circ \]

### Step 5: Substitute \( \tan 30^\circ \)
We know that \( \tan 30^\circ = \frac{1}{\sqrt{3}} \). Therefore:
\[ \sqrt{3} \tan 5^\circ \tan 15^\circ \tan 30^\circ \tan 75^\circ \tan 85^\circ = \sqrt{3} \cdot \frac{1}{\sqrt{3}} = 1 \]

### Step 6: Combine the results
Now, we can combine our results:
\[ \sin^2 25^\circ + \sin^2 65^\circ + \sqrt{3} \tan 5^\circ \tan 15^\circ \tan 30^\circ \tan 75^\circ \tan 85^\circ = 1 + 1 = 2 \]

### Final Answer
Thus, the value of the expression is:
\[ \boxed{2} \]

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