Question

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

• A. $(0,1)$
• B. $(0,\infty )$
• C. $(-1,0)$
• D. $(-\infty ,-1)$

Answer 1

The correct option is B $(0,\infty )$

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

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

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

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

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

$u=\sec x+\tan x$

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

when $x\to 0,u\to 1$

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

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