If \({\left( {{\rm{ta}}{{\rm{n}}^{ - 1}}{\rm{\;x}}} \right)^2} + {\left( {{\rm{co}}{{\rm{t}}^{ - 1}}{\rm{\;x}}} \right)^2} = \frac{{5{{\rm{\pi }}^2}}}{8},\) then x is equal to

1
–1
2
1
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0
4
2

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