Here's the question you clicked on:
Hannah_Ahn
prove the identity: \[\left( \cot x - 1 \over 1 - \tan x\right) = \left( \csc x \over \sec x \right)\]
left hand side is \[\frac{\frac{1}{b}}{\frac{1}{a}}=\frac{a}{b}\]
sorry that was right hand side
left hand side is \[\frac{\frac{a}{b}-1}{1-\frac{b}{a}}=\frac{a^2-ab}{ab-b^2}=\frac{a(a-b)}{b(a-b)}=\frac{a}{b}\] so there are the same. there is no trig here
that is an awesome way to prove it! you are amazing! thanks!!!