Question 894134
<pre>

The other tutor assumed you meant only the first quadrant. But
the tangent is positive in both the first and the third quadrants.

{{{tan(theta)=2}}}

Two cases.  {{{theta}}} is in quadrant 1, and quadrant 3

Case 1, the only one the other tutor considered):

{{{tan(theta)=y/x=2/1}}}

so take y=2, x=1

{{{r=sqrt(x^2+y^2)=sqrt(1^2+2^2)=sqrt(1+4)=sqrt(5)}}}

{{{sin(theta)=y/r=2/sqrt(5)=2sqrt(5)/5}}}

{{{cos(theta)=x/r=1/sqrt(5)=sqrt(5)/5}}}

{{{cot(theta)=x/y=1/2}}}

{{{sec(theta)=r/x=sqrt(5)/1=sqrt(5)}}}

Case 2:

{{{tan(theta)=y/x=(-2)/(-1)}}}

so take y=-2, x=-1

{{{r=sqrt(x^2+y^2)=sqrt((-1)^2+(-2)^2)=sqrt(1+4)=sqrt(5)}}}

{{{sin(theta)=y/r=(-2)/sqrt(5)=-2sqrt(5)/5}}}

{{{cos(theta)=x/r=(-1)/sqrt(5)=-sqrt(5)/5}}}

{{{cot(theta)=x/y=(-1)/(-2)=1/2}}}

{{{sec(theta)=r/x=sqrt(5)/(-1)=-sqrt(5)}}}

Edwin</pre>