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 start with the left-hand side (LHS) and simplify it step by step.

### Step 1: Write down the left-hand side
We start with the left-hand side:
\[ \text{LHS} = \tan(45^\circ - \theta) \]

### Step 2: Use the tangent subtraction formula
We apply the tangent subtraction formula:
\[ \tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B} \]
Here, \( A = 45^\circ \) and \( B = \theta \). Therefore, we have:
\[ \tan(45^\circ - \theta) = \frac{\tan 45^\circ - \tan \theta}{1 + \tan 45^\circ \tan \theta} \]

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

### Step 4: Simplify the expression
This simplifies to:
\[ \tan(45^\circ - \theta) = \frac{1 - \tan \theta}{1 + \tan \theta} \]

### Step 5: Conclusion
Since we have shown that:
\[ \tan(45^\circ - \theta) = \frac{1 - \tan \theta}{1 + \tan \theta} \]
This means that:
\[ \text{LHS} = \text{RHS} \]
Thus, we have proved the identity.