1/secx-tanx=secx tanx

Is the 1 divided by the secx alone or is it over secx - tanx together?

\frac{1}{\sec x-\tan x}=\sec x+\tan x

Multiply both sides by \sec x-\tan x and this reads

1= \sec^2 x-\tan^2 x

which is solved by direct computation

\sec^2 x-\tan^2 x = \frac{1}{\cos^2x}-\frac{\sin^2x}{\cos^2x}=\frac{1-\sin^2x}{\cos^2x}=1

The is over secx-tanx together because the is the common denominator