If A+B=left[begin{array}{cr}1 & tan frac{thet | vedclass

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(D) Given,$A+B=\left[\begin{array}{cc}1 & 	an \frac{	heta}{2} \ -	an \frac{	heta}{2} & 1\end{array}ight] \dots(i)$Taking transpose on both side

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$\left[\begin{array}{cc}0 & -2 \sqrt{3} \ 2 \sqrt{3} & 0\end{array}ight]$

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(D) Given,$A+B=\left[\begin{array}{cc}1 & 	an \frac{	heta}{2} \ -	an \frac{	heta}{2} & 1\end{array}ight] \dots(i)$Taking transpose on both sides,$A^T+B^T=\left[\begin{array}{cc}1 & -	an \frac{	heta}{2} \ 	an \frac{	heta}{2} & 1\end{array}ight]$.Since $A$ is symmetric $(A^T=A)$ and $B$ is skew-symmetric $(B^T=-B)$,we have $A-B=\left[\begin{array}{cc}1 & -	an \frac{	heta}{2} \ 	an \frac{	heta}{2} & 1\end{array}ight] \dots(ii)$.Adding $(i)$ and $(ii)$,$2A = \left[\begin{array}{cc}2 & 0 \ 0 & 2\end{array}ight] \implies A = I = \left[\begin{array}{cc}1 & 0 \ 0 & 1\end{array}ight]$.Subtracting $(ii)$ from $(i)$,$2B = \left[\begin{array}{cc}0 & 2 	an \frac{	heta}{2} \ -2 	an \frac{	heta}{2} & 0\end{array}ight] \implies B = \left[\begin{array}{cc}0 & 	an \frac{	heta}{2} \ -	an \frac{	heta}{2} & 0\end{array}ight]$.Then $A^{-1} = I^{-1} = I$ and $B^{-1} = \frac{1}{	an^2 \frac{	heta}{2}} \left[\begin{array}{cc}0 & -	an \frac{	heta}{2} \ 	an \frac{	heta}{2} & 0\end{array}ight] = \left[\begin{array}{cc}0 & -\cot \frac{	heta}{2} \ \cot \frac{	heta}{2} & 0\end{array}ight]$.Now,$A^{-1}B + AB^{-1} = B + B^{-1} = \left[\begin{array}{cc}0 & 	an \frac{	heta}{2} - \cot \frac{	heta}{2} \ -	an \frac{	heta}{2} + \cot \frac{	heta}{2} & 0\end{array}ight]$.Using $	an \frac{	heta}{2} - \cot \frac{	heta}{2} = -2 \cot 	heta$,we get $A^{-1}B + AB^{-1} = \left[\begin{array}{cc}0 & -2 \cot 	heta \ 2 \cot 	heta & 0\end{array}ight]$.At $	heta = \frac{\pi}{6}$,$\cot \frac{\pi}{6} = \sqrt{3}$.Thus,the matrix is $\left[\begin{array}{cc}0 & -2 \sqrt{3} \ 2 \sqrt{3} & 0\end{array}ight]$.

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

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