Question 1103429
For acute angles (less than 90 degrees),
you could think of trigonometric functions as trigonometric ratios.
If you draw a right triangle with {{{theta}}} as one of the angles,
the measure of the other acute angle will be {{{90-theta}}} .
{{{drawing(300,300,-0.5,4.5,-1,4,
triangle(0,0,4,0,0,3),rectangle(0,0,0.1,0.1),
locate(-0.2,0.1,A),locate(4.1,0.1,B),locate(-0.2,3.2,C)
)}}}
{{{C=180^o-B}}}
{{{tan(B)= AC / AB}}}= (opposite side) / (adjacent side)
{{{tan(180^o-B)=tan(C)=AB/AC}}} .
So,
{{{tan(180^o-B)*tan(B)=(AB/AC)*(AC / AB)=1}}}
It works for acute angle {{{B}}} from a right triangle,
and it also works for any angle {{{theta}}} ,
so {{{tan(180^o-theta)*tan(theta)=1}}} , and
{{{tan(180^o-theta)*tan(theta)*sec(theta) =1*sec(theta) =sec(theta) }}}
 
NOTE: I did not homeschool my children,
I was not brave enough for that,
but I coached my youngest through an online charter school.