Question 917709
{{{sin(theta)^2+cos(theta)^2=1}}}
{{{x^2/16+cos(theta)^2=1}}}
{{{cos(theta)^2=1-x^2/16}}}
{{{cos(theta)=0 +- sqrt(1-x^2/16)}}}
Since {{{theta}}} is in the first quadrant, {{{cos(theta)>=0}}}, 
so,
{{{cos(theta)=sqrt(1-x^2/16)}}}
and
{{{tan(theta)=sin(theta)/cos(theta)}}}
{{{tan(theta)=(x/16)/(sqrt(1-x^2/16))}}}
{{{tan(theta)=x/(4*sqrt(16-x^2)))}}}