Question

prove the identity:
\[\left( \cot x - 1 \over 1 - \tan x\right) = \left( \csc x \over \sec x \right)\]

Answer 1

left hand side is
\[\frac{\frac{1}{b}}{\frac{1}{a}}=\frac{a}{b}\]

Answer 2

sorry that was right hand side

Answer 3

it's okay (-:

Answer 4

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

Answer 5

that is an awesome way to prove it! you are amazing! thanks!!!