If `[(1,-tan theta),(tan theta,1)][(1,tan theta),(-tan theta,1)]^(-1)=[(a,-b),(b,a)],` then

Options

A. `a= cos 2 theta`
B. `a=1`
C. `b= sin 2 theta`
D. `b=-1`

Explanation

To solve the problem, we need to find the values of \( a \) and \( b \) in the equation:

\[ \left[ \begin{pmatrix} 1 & -\tan \theta \\ \tan \theta & 1 \end{pmatrix} \cdot \begin{pmatrix} 1 & \tan \theta \\ -\tan \theta & 1 \end{pmatrix} \right]^{-1} = \begin{pmatrix} a & -b \\ b & a \end{pmatrix} \]

### Step 1: Calculate the product of the two matrices

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