If `tan^(2) (( pi )/( 4) + ( theta )/( 2)) = - ( a )/( b)`, then

Options

A. `sin theta = ( a-b)/( a+b)`
B. `sin theta = ( a+ b)/( a-b)`
C. `cos theta = ( a+b)/( a-b)`
D. `tan theta = ( a-b)/( a+b)`

Answer

To solve the equation \( \tan^2 \left( \frac{\pi}{4} + \frac{\theta}{2} \right) = -\frac{a}{b} \), we will follow these steps:

### Step 1: Use the tangent addition formula
We start with the tangent addition formula:
\[ \tan\left(A + B\right) = \frac{\tan A + \tan B}{1 - \tan A \tan B} \]
Let \( A = \frac{\pi}{4} \) and \( B = \frac{\theta}{2} \). Thus, we have:
\[ \tan\left(\frac{\pi}{4} + \frac{\theta}{2}\right) = \frac{\tan\left(\frac{\pi}{4}\right) + \tan\left(\frac{\theta}{2}\right)}{1 - \tan\left(\frac{\pi}{4}\right) \tan\left(\frac{\theta}{2}\right)} \]

### Step 2: Substitute the value of \(\tan\left(\frac{\pi}{4}\right)\)
Since \( \tan\left(\frac{\pi}{4}\right) = 1 \), we can substitute this into our equation:
\[ \tan\left(\frac{\pi}{4} + \frac{\theta}{2}\right) = \frac{1 + \tan\left(\frac{\theta}{2}\right)}{1 - \tan\left(\frac{\theta}{2}\right)} \]

### Step 3: Square both sides
Now we square both sides to find \( \tan^2\left(\frac{\pi}{4} + \frac{\theta}{2}\right) \):
\[ \tan^2\left(\frac{\pi}{4} + \frac{\theta}{2}\right) = \left(\frac{1 + \tan\left(\frac{\theta}{2}\right)}{1 - \tan\left(\frac{\theta}{2}\right)}\right)^2 \]

### Step 4: Set the equation equal to \(-\frac{a}{b}\)
From the problem statement, we have:
\[ \left(\frac{1 + \tan\left(\frac{\theta}{2}\right)}{1 - \tan\left(\frac{\theta}{2}\right)}\right)^2 = -\frac{a}{b} \]

### Step 5: Analyze the equation
Since the left-hand side is a square, it is always non-negative. Therefore, for the equation to hold, we must have:
\[ -\frac{a}{b} \leq 0 \implies a \geq 0 \]
This implies that \( a \) must be non-negative.

### Step 6: Conclusion
Thus, the equation \( \tan^2\left(\frac{\pi}{4} + \frac{\theta}{2}\right) = -\frac{a}{b} \) can only hold true if \( a \) is non-negative and \( b \) is positive.