$AB=I$
$B=A^{-1}$
$\;\;\;=\large\frac{1}{|A|}$$adj.A$
$\;\;\;=\large\frac{1}{1+\tan^2\large\frac{\theta}{2}}$$\begin{bmatrix}1&\tan\large\frac{\theta}{2}\\-\tan\large\frac{\theta}{2}& 1\end{bmatrix}$
$\;\;\;=(\cos^2\large\frac{\theta}{2})$$ A'$
Hence (B) is the correct answer.