Prove that `(1+tanA)^(2)+(1-tanA)^(2)=2sec^(2)A`.

Answer

To prove that \((1 + \tan A)^2 + (1 - \tan A)^2 = 2 \sec^2 A\), we will start by expanding the left-hand side (LHS) and simplifying it step by step.

### Step 1: Expand the LHS
We start with the expression:
\[ (1 + \tan A)^2 + (1 - \tan A)^2 \]
Using the formula \((a + b)^2 = a^2 + 2ab + b^2\), we can expand both squares:
\[ (1 + \tan A)^2 = 1^2 + 2 \cdot 1 \cdot \tan A + \tan^2 A = 1 + 2\tan A + \tan^2 A \]
\[ (1 - \tan A)^2 = 1^2 - 2 \cdot 1 \cdot \tan A + \tan^2 A = 1 - 2\tan A + \tan^2 A \]
Now, adding these two expansions together:
\[ (1 + \tan A)^2 + (1 - \tan A)^2 = (1 + 2\tan A + \tan^2 A) + (1 - 2\tan A + \tan^2 A) \]

### Step 2: Combine Like Terms
Combining the terms from the expansion:
\[ = 1 + 2\tan A + \tan^2 A + 1 - 2\tan A + \tan^2 A \]
Notice that \(2\tan A\) and \(-2\tan A\) cancel each other out:
\[ = 1 + 1 + \tan^2 A + \tan^2 A = 2 + 2\tan^2 A \]

### Step 3: Factor Out the Common Term
Now we can factor out the common term \(2\):
\[ = 2(1 + \tan^2 A) \]

### Step 4: Use the Trigonometric Identity
We know from the Pythagorean identity that:
\[ 1 + \tan^2 A = \sec^2 A \]
Substituting this identity into our expression gives:
\[ = 2 \sec^2 A \]

### Conclusion
Thus, we have shown that:
\[ (1 + \tan A)^2 + (1 - \tan A)^2 = 2 \sec^2 A \]
This completes the proof.