Prove: `tan (45^(@)+ theta ) = (1 + tan theta)/( 1- tan theta )`

Answer

To prove the identity \( \tan(45^\circ + \theta) = \frac{1 + \tan \theta}{1 - \tan \theta} \), we will use the tangent addition formula.

### Step-by-Step Solution:

1. **Start with the Left-Hand Side (LHS)**:
\[ \tan(45^\circ + \theta) \]

2. **Apply the Tangent Addition Formula**:
The formula for the tangent of a sum is:
\[ \tan(a + b) = \frac{\tan a + \tan b}{1 - \tan a \tan b} \]
Here, let \( a = 45^\circ \) and \( b = \theta \). Thus, we have:
\[ \tan(45^\circ + \theta) = \frac{\tan 45^\circ + \tan \theta}{1 - \tan 45^\circ \tan \theta} \]

3. **Substitute the Value of \( \tan 45^\circ \)**:
We know that \( \tan 45^\circ = 1 \). Substituting this value into the equation gives:
\[ \tan(45^\circ + \theta) = \frac{1 + \tan \theta}{1 - 1 \cdot \tan \theta} \]

4. **Simplify the Denominator**:
The denominator simplifies to:
\[ 1 - \tan \theta \]
Therefore, we have:
\[ \tan(45^\circ + \theta) = \frac{1 + \tan \theta}{1 - \tan \theta} \]

5. **Conclusion**:
Since we have shown that the left-hand side equals the right-hand side, we conclude:
\[ \tan(45^\circ + \theta) = \frac{1 + \tan \theta}{1 - \tan \theta} \]
Hence, the identity is proved.