If A = \(\begin{bmatrix}\tan {\alpha }&-\sin {\alpha }\\\sin {\alpha }&\tan {\alpha }\end{bmatrix}\) and \(A+ { A }^{ T } \) = \(\frac{2}{\sqrt { 3 }}\) I, then the value of \(\alpha \) is

\(\frac{\Pi }{6}\)

A = \(\begin{bmatrix}\tan {\alpha }&-\sin {\alpha }\\\sin {\alpha }&\tan {\alpha }\end{bmatrix}\)

\( { A }^{ T } \) = \(\begin{bmatrix}\tan {\alpha }&\sin {\alpha }\\-\sin {\alpha }&\tan{\alpha } \end{bmatrix}\)

\(A+ { A }^{ T } \) = \(\begin{bmatrix}\tan {\alpha }&-\sin {\alpha }\\\sin {\alpha }&\tan {\alpha }\end{bmatrix}\) + \(\begin{bmatrix}\tan {\alpha }&\sin {\alpha }\\-\sin {\alpha }&\tan{\alpha }\end{bmatrix}\)= \(\frac{2}{\sqrt { 3 }}\) \(\begin{bmatrix}1&0\\0&1\end{bmatrix}\)

\(\begin{bmatrix}2\tan{\alpha }&0\\0&2\tan {\alpha }\end{bmatrix}\) = \(\frac{2}{\sqrt { 3 }}\) \(\begin{bmatrix}1&0\\0&1\end{bmatrix}\)

\(2\tan {\alpha }=\frac{2}{\sqrt { 3 }}\)

\(\tan {\alpha }=\frac{1}{\sqrt { 3 }}\)

\(\tan {\alpha } = \tan {\frac{\Pi }{6}}\)

Thus , \(\alpha =\frac{\Pi }{6}\)