Range of $f(x)=\frac{secx+tanx-1}{tanx-secx+1}:xϵ(0,\frac{\pi }{2})$ is -

$f(x)=\frac{\sec x+\tan x-1}{\tan x-\sec x+1}$

$f(x)=\frac{\sec x+\tan x-1}{\frac{1}{\sec x+\tan x}+1}$

substitute $\sec x+\tan x=u$

$f(x)=\frac{u-1}{\frac{1}{u}+1}$

$f(x)=\frac{u(u-1)}{u+1}$

$u=\sec x+\tan x$

$u^{\prime }=\sec x\tan x+{\sec }^{2}x>0$  in  $\left(0,\frac{\pi }{2}\right)$

when  $x\to 0,u\to 1$

when  $x\to \frac{\pi }{2},u\to \infty$

Therefore the range of $f(x)=(0,\infty )$