If `t a n(pi//4+theta)+t a n(pi//4-theta)=a ,` then `t a n^2(pi//4+theta)+t a n^2(pi//4-theta)` = (a) `a^2+1` (b) `a^2+2` (c) `a^2-2` (d) none of these

Explanation

To solve the problem, we need to find the value of \( \tan^2\left(\frac{\pi}{4} + \theta\right) + \tan^2\left(\frac{\pi}{4} - \theta\right) \) given that \( \tan\left(\frac{\pi}{4} + \theta\right) + \tan\left(\frac{\pi}{4} - \theta\right) = a \).

### Step-by-Step Solution:

1. **Use the tangent addition and subtraction formulas**:
We know that:
\[
\tan\left(\frac{\pi}{4} + \theta\right) = \frac{1 + \tan\theta}{1 - \tan\theta}
...