If \(\mathrm{A}+\mathrm{B}=\left[\begin{array}{cr}1 & \tan \frac{\theta}{2} \\ -\tan \frac{\theta}{2} & 1\end{array}\right]\) where A is symmetric and B is skew-symmetric matrix, then the matrix \(\left(A^{-1} B+A B^{-1}\right)\) at \(\theta=\frac{\pi}{6}\) is given by

Given, \(A+B=\left[\begin{array}{cc}1 & \tan \frac{\theta}{2} \\ -\tan \frac{\theta}{2} & 1\end{array}\right]\)...(i)
\(A^T+B^T=\left[\begin{array}{cc}
1 & -\tan \frac{\theta}{2} \\
\tan \frac{\theta}{2} & 1
\end{array}\right]\)
Since \(A\) is symmetric and \(B\) is skew symmetric
\(\begin{aligned}
& A^T=A, B^T=-B \\
\therefore & A-B=\left[\begin{array}{cc}
1 & -\tan \frac{\theta}{2} \\
\tan \frac{\theta}{2} & 1
\end{array}\right]...(ii)
\end{aligned}\)
Adding (i) and (ii), we get
\(\begin{aligned}
2 A & =\left[\begin{array}{cc}
1 & \tan \frac{\theta}{2} \\
-\tan \frac{\theta}{2} & 1
\end{array}\right]+\left[\begin{array}{cc}
1 & -\tan \frac{\theta}{2} \\
\tan \frac{\theta}{2} & 1
\end{array}\right] \\
2 A & =\left[\begin{array}{ll}
2 & 0 \\
0 & 2
\end{array}\right] \\
\therefore \quad A & =\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]
\end{aligned}\)
Now, subtracting (ii) from (i), we get
\(\begin{aligned}
& 2 B=\left[\begin{array}{cc}
1 & \tan \frac{\theta}{2} \\
-\tan \frac{\theta}{2} & 1
\end{array}\right]-\left[\begin{array}{cc}
1 & -\tan \frac{\theta}{2} \\
\tan \frac{\theta}{2} & 1
\end{array}\right] \\
& 2 B=\left[\begin{array}{cc}
0 & 2 \tan \frac{\theta}{2} \\
-2 \tan \frac{\theta}{2} & 0
\end{array}\right] \\
& \therefore B=\left[\begin{array}{cc}
0 & \tan \frac{\theta}{2} \\
-\tan \frac{\theta}{2} & 0
\end{array}\right]
\end{aligned}\)
\(\begin{array}{ll}
\therefore & B^{-1}=\frac{1}{\tan ^2 \frac{\theta}{2}}\left[\begin{array}{cc}
0 & -\tan \frac{\theta}{2} \\
\tan \frac{\theta}{2} & 0
\end{array}\right] \\
\therefore & B^{-1}=\left[\begin{array}{cc}
0 & \frac{-1}{\tan \frac{\theta}{2}} \\
\frac{1}{\tan \frac{\theta}{2}} & 0
\end{array}\right]
\end{array}\)
\(\therefore \quad\) Now, \(\left(\mathrm{A}^{-1} \mathrm{~B}+\mathrm{AB}^{-1}\right)\)
\(=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]\left[\begin{array}{cc}0 & \tan \frac{\theta}{2} \\ -\tan \frac{\theta}{2} & 0\end{array}\right]+\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]\left[\begin{array}{cc}0 & \frac{-1}{\tan \frac{\theta}{2}} \\ \frac{1}{\tan \frac{\theta}{2}} & 0\end{array}\right]\)
\(=\left[\begin{array}{cc}0 & \tan \frac{\theta}{2} \\ -\tan \frac{\theta}{2} & 0\end{array}\right]+\left[\begin{array}{cc}0 & \frac{-1}{\tan \frac{\theta}{2}} \\ \frac{1}{\tan \frac{\theta}{2}} & 0\end{array}\right]\)
\(\begin{aligned} & =\left[\begin{array}{cc}0 & \frac{\tan ^2 \frac{\theta}{2}-1}{\tan \frac{\theta}{2}} \\ \frac{1-\tan ^2 \frac{\theta}{2}}{\tan \frac{\theta}{2}} & 0\end{array}\right] \\ & =2\left[\begin{array}{cc}0 & (-1) \frac{1-\tan ^2 \frac{\theta}{2}}{2 \tan \frac{\theta}{2}} \\ \frac{1-\tan ^2 \frac{\theta}{2}}{2 \tan \frac{\theta}{2}} & 0\end{array}\right]\end{aligned}\)
\(\begin{aligned} & \therefore=2\left[\begin{array}{cc}0 & \frac{-1}{\tan \theta} \\ \frac{1}{\tan \theta} & 0\end{array}\right] \\ & \therefore \quad\left(\mathrm{A}^{-1} \mathrm{~B}+\mathrm{AB}^{-1}\right) \text { at } \theta=\frac{\pi}{6}\end{aligned}\)
\(\begin{aligned} & =2\left[\begin{array}{cc}0 & \frac{-1}{\tan \frac{\pi}{6}} \\ \frac{1}{\tan \frac{\pi}{6}} & 0\end{array}\right] \\ & =2\left[\begin{array}{cc}0 & -\sqrt{3} \\ \sqrt{3} & 0\end{array}\right] \\ & =\left[\begin{array}{cc}0 & -2 \sqrt{3} \\ 2 \sqrt{3} & 0\end{array}\right]\end{aligned}\)