If $${\text{sin}}\left( {{{60}^ \circ } - \theta } \right)$$   = $${\text{cos}}\left( {\psi - {{30}^ \circ }} \right),$$   then the value of $${\text{tan}}\left( {\psi - \theta } \right)$$   is (assume that $$\theta $$ and $$\psi $$ are both positive acute angles with$$\theta < {60^ \circ }$$ and $$\psi > {30^ \circ }$$  ) ?

$$\eqalign{ & {\text{sin}}\left( {{{60}^ \circ } - \theta } \right) = {\text{cos}}\left( {\psi - {{30}^ \circ }} \right) \cr & \Rightarrow \left( {{{60}^ \circ } - \theta } \right) + \left( {\psi - {{30}^ \circ }} \right) = {90^ \circ } \cr & \left[ {{\text{If sin A}} = {\text{cos B}}\,{\text{then, A}} + {\text{B}} = {{90}^ \circ }} \right] \cr & \Rightarrow \left( {\psi - \theta } \right) = {90^ \circ } - {30^ \circ } \cr & \Rightarrow \left( {\psi - \theta } \right) = {60^ \circ } \cr & \Rightarrow \tan \left( {\psi - \theta } \right) = \tan {60^ \circ } \cr & \Rightarrow \tan {60^ \circ } = \sqrt 3 \cr} $$