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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$$\theta {30^ \circ }$$  ) ?

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$$\theta {30^ \circ }$$  ) ? Correct Answer $$\sqrt 3 $$

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

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Evaluate the following:
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