How do you prove that the sin over one minus the cosine minus one plus the cosine over the sine equals zero?

Answer 1

Multiply both sides by sin(1-cos) and you lose the denominators and get (sin squared) minus 1+cos times 1-cos.

Then multiply out (i.e. expand) 1+cos times 1-cos, which will of course give the difference of two squares: 1 - (cos squared). (because the cross terms cancel out.)

(This is diff of 2 squares because 1 is the square of 1.)

And so you get (sin squared) - (1 - (cos squared)) = (sin squared) + (cos squared) - 1.

Then from basic trig we know that (sin squared) + (cos squared) = 1, so this is 0.