`[[1, -tan((theta)/(2))], [tan((theta)/(2)), 1]][[1, tan((theta)/(2))], [-tan((theta)/(2)), 1]]^(-1)=`

Options

A. `[[1, 0], [0, 1]]`
B. `[[0, 0], [0, 0]]`
C. `[[costheta, sintheta], [sintheta, costheta]]`
D. `[[costheta, -sintheta], [sintheta, costheta]]`

Answer

To solve the given matrix equation

\[ \begin{bmatrix} 1 & -\tan\left(\frac{\theta}{2}\right) \\ \tan\left(\frac{\theta}{2}\right) & 1 \end{bmatrix} \begin{bmatrix} 1 & \tan\left(\frac{\theta}{2}\right) \\ -\tan\left(\frac{\theta}{2}\right) & 1 \end{bmatrix}^{-1} \]

we will follow these steps:

### Step 1: Define Variables
Let \( x = \tan\left(\frac{\theta}{2}\right) \). Then, we can rewrite our matrices as:

\[ A = \begin{bmatrix} 1 & -x \\ x & 1 \end{bmatrix} \]
and
\[ B = \begin{bmatrix} 1 & x \\ -x & 1 \end{bmatrix} \]

### Step 2: Find the Inverse of Matrix B
The inverse of a \( 2 \times 2 \) matrix \( B = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \) is given by:

\[ B^{-1} = \frac{1}{ad - bc} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix} \]

For our matrix \( B \):

- \( a = 1 \), \( b = x \), \( c = -x \), \( d = 1 \)
- The determinant \( \text{det}(B) = ad - bc = 1 \cdot 1 - (-x) \cdot x = 1 + x^2 \)

Thus, the inverse of \( B \) is:

\[ B^{-1} = \frac{1}{1 + x^2} \begin{bmatrix} 1 & -x \\ x & 1 \end{bmatrix} \]

### Step 3: Multiply Matrices A and \( B^{-1} \)
Now we need to compute \( A \cdot B^{-1} \):

\[ A \cdot B^{-1} = \begin{bmatrix} 1 & -x \\ x & 1 \end{bmatrix} \cdot \frac{1}{1 + x^2} \begin{bmatrix} 1 & -x \\ x & 1 \end{bmatrix} \]

Calculating the product:

1. First row, first column:
\[ 1 \cdot 1 + (-x) \cdot x = 1 - x^2 \]

2. First row, second column:
\[ 1 \cdot (-x) + (-x) \cdot 1 = -x - x = -2x \]

3. Second row, first column:
\[ x \cdot 1 + 1 \cdot x = x + x = 2x \]

4. Second row, second column:
\[ x \cdot (-x) + 1 \cdot 1 = -x^2 + 1 = 1 - x^2 \]

Thus, we have:

\[ A \cdot B^{-1} = \frac{1}{1 + x^2} \begin{bmatrix} 1 - x^2 & -2x \\ 2x & 1 - x^2 \end{bmatrix} \]

### Step 4: Substitute Back for \( x \)
Now we substitute back \( x = \tan\left(\frac{\theta}{2}\right) \):

\[ = \frac{1}{1 + \tan^2\left(\frac{\theta}{2}\right)} \begin{bmatrix} 1 - \tan^2\left(\frac{\theta}{2}\right) & -2\tan\left(\frac{\theta}{2}\right) \\ 2\tan\left(\frac{\theta}{2}\right) & 1 - \tan^2\left(\frac{\theta}{2}\right) \end{bmatrix} \]

### Step 5: Simplify Using Trigonometric Identities
Using the identity \( 1 + \tan^2\left(\frac{\theta}{2}\right) = \sec^2\left(\frac{\theta}{2}\right) \) and \( 1 - \tan^2\left(\frac{\theta}{2}\right) = \cos\theta \):

\[ = \cos\theta \begin{bmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{bmatrix} \]

### Final Answer
The final answer is:

\[ \begin{bmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{bmatrix} \]