Prove that
`tan ((pi )/(4) + (theta )/(2) ) + tan ((pi )/(4) - (theta)/(2)) = 2 sec theta.`

Answer

To prove that

\[ \tan\left(\frac{\pi}{4} + \frac{\theta}{2}\right) + \tan\left(\frac{\pi}{4} - \frac{\theta}{2}\right) = 2 \sec \theta, \]

we will start by using the tangent addition and subtraction formulas.

### Step 1: Write the Left-Hand Side (LHS)

Let’s denote the left-hand side (LHS) as:

\[ LHS = \tan\left(\frac{\pi}{4} + \frac{\theta}{2}\right) + \tan\left(\frac{\pi}{4} - \frac{\theta}{2}\right). \]

### Step 2: Apply the Tangent Addition Formula

Using the tangent addition formula:

\[ \tan(a + b) = \frac{\tan a + \tan b}{1 - \tan a \tan b}, \]

we can substitute \( a = \frac{\pi}{4} \) and \( b = \frac{\theta}{2} \) for the first term:

\[ \tan\left(\frac{\pi}{4} + \frac{\theta}{2}\right) = \frac{\tan\left(\frac{\pi}{4}\right) + \tan\left(\frac{\theta}{2}\right)}{1 - \tan\left(\frac{\pi}{4}\right) \tan\left(\frac{\theta}{2}\right)}. \]

Since \( \tan\left(\frac{\pi}{4}\right) = 1 \):

\[ \tan\left(\frac{\pi}{4} + \frac{\theta}{2}\right) = \frac{1 + \tan\left(\frac{\theta}{2}\right)}{1 - \tan\left(\frac{\theta}{2}\right)}. \]

### Step 3: Apply the Tangent Subtraction Formula

Now for the second term, using the tangent subtraction formula:

\[ \tan\left(\frac{\pi}{4} - \frac{\theta}{2}\right) = \frac{\tan\left(\frac{\pi}{4}\right) - \tan\left(\frac{\theta}{2}\right)}{1 + \tan\left(\frac{\pi}{4}\right) \tan\left(\frac{\theta}{2}\right)}. \]

Again substituting \( \tan\left(\frac{\pi}{4}\right) = 1 \):

\[ \tan\left(\frac{\pi}{4} - \frac{\theta}{2}\right) = \frac{1 - \tan\left(\frac{\theta}{2}\right)}{1 + \tan\left(\frac{\theta}{2}\right)}. \]

### Step 4: Combine Both Terms

Now substituting both terms back into the LHS:

\[ LHS = \frac{1 + \tan\left(\frac{\theta}{2}\right)}{1 - \tan\left(\frac{\theta}{2}\right)} + \frac{1 - \tan\left(\frac{\theta}{2}\right)}{1 + \tan\left(\frac{\theta}{2}\right)}. \]

### Step 5: Find a Common Denominator

To combine these fractions, we find a common denominator:

\[ LHS = \frac{(1 + \tan\left(\frac{\theta}{2}\right))^2 + (1 - \tan\left(\frac{\theta}{2}\right))^2}{(1 - \tan\left(\frac{\theta}{2}\right))(1 + \tan\left(\frac{\theta}{2}\right))}. \]

### Step 6: Simplify the Numerator

Expanding the numerator:

\[ (1 + \tan\left(\frac{\theta}{2}\right))^2 = 1 + 2\tan\left(\frac{\theta}{2}\right) + \tan^2\left(\frac{\theta}{2}\right), \]
\[ (1 - \tan\left(\frac{\theta}{2}\right))^2 = 1 - 2\tan\left(\frac{\theta}{2}\right) + \tan^2\left(\frac{\theta}{2}\right). \]

Adding these gives:

\[ 2 + 2\tan^2\left(\frac{\theta}{2}\right). \]

### Step 7: Write the LHS

Thus, we have:

\[ LHS = \frac{2 + 2\tan^2\left(\frac{\theta}{2}\right)}{1 - \tan^2\left(\frac{\theta}{2}\right)}. \]

### Step 8: Factor Out 2

Factoring out 2 from the numerator:

\[ LHS = 2 \cdot \frac{1 + \tan^2\left(\frac{\theta}{2}\right)}{1 - \tan^2\left(\frac{\theta}{2}\right)}. \]

### Step 9: Use the Identity

Using the identity \( 1 + \tan^2 x = \sec^2 x \):

\[ LHS = 2 \cdot \frac{\sec^2\left(\frac{\theta}{2}\right)}{\cos\left(\theta\right)}. \]

### Step 10: Use Cosine Double Angle Identity

Using the identity \( \cos\theta = \frac{1 - \tan^2\left(\frac{\theta}{2}\right)}{1 + \tan^2\left(\frac{\theta}{2}\right)} \):

\[ LHS = 2 \sec\theta. \]

### Conclusion

Thus, we have shown that:

\[ LHS = 2 \sec \theta, \]

which proves that

\[ \tan\left(\frac{\pi}{4} + \frac{\theta}{2}\right) + \tan\left(\frac{\pi}{4} - \frac{\theta}{2}\right) = 2 \sec \theta. \]