Question

Another one that I need to prove.

Answer 1

\[(\sec x -\tan x)^2=(1-\sin x)/(1+\sin x)\]

Answer 2

What have you tried?

Answer 3

I multiplied the right side by the conjugate but I got stuck. I ended up with \[(1-2\sin x+\sin^2x)/(\sin x+1)\]

Answer 4

Should I try the left side instead?

Answer 5

\[ ( \sec x -\tan x)^2 = \frac 1{\cos^2x} + \frac {\sin^2 x}{\cos^2x} - \frac{2 \sin x}{\cos^2 x} \]\[ = \frac{1+ \sin^2x -2 \sin x}{1-\sin^2 x} = \frac{1-\sin x}{1+\sin x}\]