Find `theta ` (in degree) satisfying the equation,`tan 15^@ *tan 25^@* tan 35^@ = tan theta,` where `theta in (0,45^@)`

Answer

To solve the equation \( \tan 15^\circ \cdot \tan 25^\circ \cdot \tan 35^\circ = \tan \theta \), where \( \theta \) is in the interval \( (0, 45^\circ) \), we can use a known trigonometric identity.

### Step-by-step Solution:

1. **Use the Identity**: We will use the identity for the product of tangents:
\[ \tan x \cdot \tan(60^\circ - x) \cdot \tan(60^\circ + x) = \tan(3x) \]
Here, we can set \( x = 5^\circ \). Thus,
\[ \tan 5^\circ \cdot \tan(60^\circ - 5^\circ) \cdot \tan(60^\circ + 5^\circ) = \tan(15^\circ) \cdot \tan(25^\circ) \cdot \tan(35^\circ) \]

2. **Calculate the Angles**: Plugging in \( x = 5^\circ \):
\[ \tan 5^\circ \cdot \tan 55^\circ \cdot \tan 65^\circ = \tan(15^\circ) \cdot \tan(25^\circ) \cdot \tan(35^\circ) \]
Since \( \tan(55^\circ) = \cot(35^\circ) \) and \( \tan(65^\circ) = \cot(25^\circ) \), we can rewrite the left-hand side:
\[ \tan 5^\circ \cdot \cot 35^\circ \cdot \cot 25^\circ \]

3. **Simplify the Equation**: The equation becomes:
\[ \tan 5^\circ = \tan 15^\circ \cdot \tan 25^\circ \cdot \tan 35^\circ \]

4. **Relate to \(\tan \theta\)**: From the equation, we can see that:
\[ \tan \theta = \tan 5^\circ \]

5. **Find \(\theta\)**: Since \( \theta \) must be in the interval \( (0, 45^\circ) \), we conclude:
\[ \theta = 5^\circ \]

### Final Answer:
Thus, the value of \( \theta \) satisfying the equation is:
\[ \theta = 5^\circ \]