Question 1128743
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1.  If   {{{cos(theta/2)}}} = a,  then  {{{cos(theta)}}} = {{{2*cos^2(theta/2)-1}}} = {{{2*a^2-1}}}.        (1)



2.  If  {{{sin(2*theta)}}} = b,  then  {{{2*sin(theta)*cos(theta)}}} = b,

    hence,  {{{sin(theta)}}} = {{{b/(2*cos(theta))}}} = (substitute from (1)) = {{{b/(2*(2*a^2-1))}}}.



3.  Then  {{{tan(theta)}}} = {{{sin(theta)/cos(theta)}}} = {{{b/(2*(2*a^2-1))}}} : {{{(2a^2-1))}}} = {{{b/(2*(2a^2-1)^2)}}}.        <U>ANSWER</U> 
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Solved.