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# If $A=\begin{bmatrix}1&\tan\large\frac{\theta}{2}\\-\tan\large\frac{\theta}{2}&1\end{bmatrix}$ and $AB=I$ then $B$=

$\begin{array}{1 1}(A)\;(\cos^2\large\frac{\theta}{2})\normalsize A&(B)\;(\cos^2\large\frac{\theta}{2})\normalsize A'\\(C)\;(\cos^2\large\frac{\theta}{2})\normalsize I&(D)\;\text{None of these}\end{array}$

$AB=I$
$B=A^{-1}$
$\;\;\;=\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.