Prove that (1 + tan A - Sec A)`xx` (1 + tan A + sec A) = 2 tan

Answer

To prove that \((1 + \tan A - \sec A)(1 + \tan A + \sec A) = 2 \tan A\), we will start by simplifying the left-hand side (LHS) of the equation.

### Step 1: Expand the left-hand side
We can recognize that the expression \((1 + \tan A - \sec A)(1 + \tan A + \sec A)\) is in the form of \( (A - B)(A + B) = A^2 - B^2 \). Here, let:
- \(A = 1 + \tan A\)
- \(B = \sec A\)

Thus, we can rewrite the LHS as:
\[ LHS = (1 + \tan A)^2 - (\sec A)^2 \]

### Step 2: Calculate \((1 + \tan A)^2\)
Now, we will calculate \((1 + \tan A)^2\):
\[ (1 + \tan A)^2 = 1^2 + 2 \cdot 1 \cdot \tan A + \tan^2 A = 1 + 2 \tan A + \tan^2 A \]

### Step 3: Substitute \(\sec^2 A\)
Using the identity \(\sec^2 A = 1 + \tan^2 A\), we can substitute \(\sec^2 A\) into our expression:
\[ LHS = (1 + 2 \tan A + \tan^2 A) - (1 + \tan^2 A) \]

### Step 4: Simplify the expression
Now, simplify the expression:
\[ LHS = 1 + 2 \tan A + \tan^2 A - 1 - \tan^2 A \]
The \(1\) and \(-1\) cancel out, and \(\tan^2 A\) also cancels out:
\[ LHS = 2 \tan A \]

### Step 5: Conclusion
Thus, we have shown that:
\[ (1 + \tan A - \sec A)(1 + \tan A + \sec A) = 2 \tan A \]
This proves that the left-hand side is equal to the right-hand side.