Question 902998
{{{sin^2(theta)+cos^2(theta)=1}}}
{{{25/49+cos^2(theta)=1}}}
{{{cos^2(theta)=49/49-25/49}}}
{{{cos^2(theta)=24/49}}}
{{{cos(theta)=sqrt(24)/7}}}
{{{cos(theta)=(2/7)sqrt(6)}}}
So then,
{{{tan(theta)=sin(theta)/cos(theta)}}}
{{{tan(theta)=(-(5/7))/((2/7)sqrt(6))}}}
{{{tan(theta)=(-5/2)(1/sqrt(6))}}}
{{{tan(theta)=(-5/2)(sqrt(6)/6)}}}
{{{tan(theta)=-(5/12)sqrt(6)}}}
The co-functions are just the inverses of these values.