Question 888933
{{{3*cot(theta)=2}}}
{{{cot(theta)=2/3}}}
{{{cos(theta)/sin(theta)=2/3}}}
{{{cos(theta)=(2/3)sin(theta)}}}
{{{cos^2(theta)=(4/9)sin^2(theta)}}}
but
{{{cos^2(theta)+sin^2(theta)=1}}}
Substituting,
{{{(4/9)sin^2(theta)+sin^2(theta)=1}}}
{{{(13/9)sin^2(theta)=1}}}
{{{sin^2(theta)=9/13}}}
{{{sin(theta)=3/sqrt(13)}}}
{{{cos(theta)=2/sqrt(13)}}}
So then,
{{{4sin(theta)-3cos(theta)=(1/sqrt(13))(4(3)-3(2))=6/sqrt(13)}}}
{{{2sin(theta)+6cos(theta)=(1/sqrt(13))(2(3)+6(2))=18/sqrt(13)}}}
and
{{{(4sin(theta)-3cos(theta))/(2sin(theta)+6cos(theta))=6/18}}}
{{{(4sin(theta)-3cos(theta))/(2sin(theta)+6cos(theta))=1/3}}}