Question 1065128
For a right triangle with an angle measuring {{{beta}}} ,
{{{tan(beta)}}}= opposite leg/adjacent leg.
If we draw such a triangle with legs measuring {{{3}}} and {{{1}}} on angle {{{beta}}} ,
it would look like this:
{{{drawing(300,300,-0.5,4.5,-0.5,4.5,
red(line(0,0,1.45,4.35)),red(arrow(0,0,1.45,4.35)),
red(line(0,0,1.45sqrt(10),0)),red(arrow(0,0,1.45sqrt(10),0)),
red(arc(0,0,8.1,8.1,-71.57,0)),locate(3,3,red(beta)),
triangle(0,0,1,0,1,3),rectangle(1,0,0.9,0.1),
locate(0.45,0.25,1),locate(0.85,1.6,3),
locate(0.4,1.2,x)
)}}} {{{x^2=1^2+3^2=1+3}}} <---> {{{x=sqrt(10)}}} as per the Pythagorean theorem.
For that triangle, we can calculate:
{{{sin(beta)=3/sqrt(10)=3sqrt(10)/10}}}
{{{cos(beta)=1/sqrt(10)=sqrt(10)/10}}}
{{{cot(beta)=1/tan(beta)=1/3}}}
{{{sec(beta)=1/cos(beta)}}}{{{"="}}}{{{1/((1/sqrt(10)))}}}{{{"="}}}{{{sqrt(10)}}}
{{{csc(beta)=1/sin(beta)}}}{{{"="}}}{{{1/((3/sqrt(10)))}}}{{{"="}}}{{{sqrt(10)/3}}}