Let $f : \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \to R$ be given by $f(x) = (\log(\sec x + \tan x))^{3}$. Then

$f(x)=(\log(\sec x+\tan x))^{3}$

$\text{Domain: } x\in\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)$

$\text{Check: odd / even, one-one, onto}$

$\sec x+\tan x=\dfrac{1}{\cos x}+\dfrac{\sin x}{\cos x}=\dfrac{1+\sin x}{\cos x}$

$x\in\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)\Rightarrow \cos x>0,\ 1+\sin x>0$

$\Rightarrow \sec x+\tan x>0$

$(\sec x+\tan x)(\sec x-\tan x)=\sec^{2}x-\tan^{2}x=1$

$\Rightarrow \sec x-\tan x=\dfrac{1}{\sec x+\tan x}$

$g(x)=\log(\sec x+\tan x)$

$g(-x)=\log(\sec x-\tan x)=\log\dfrac{1}{\sec x+\tan x}=-g(x)$

$\Rightarrow g \text{ is odd}$

$f(x)=(g(x))^{3},\quad f(-x)=(-g(x))^{3}=-f(x)$

$\Rightarrow f \text{ is odd} \Rightarrow (1) \text{ true},\ (4) \text{ false}$

$g'(x)=\dfrac{\sec x\tan x+\sec^{2}x}{\sec x+\tan x}=\dfrac{\sec x(\sec x+\tan x)}{\sec x+\tan x}=\sec x$

$\sec x>0 \Rightarrow g \text{ increasing} \Rightarrow f=g^{3} \text{ increasing}$

$\Rightarrow f \text{ one-one} \Rightarrow (2) \text{ true}$

$x\to \dfrac{\pi}{2}^{-}:\ \sec x+\tan x\to+\infty \Rightarrow g\to+\infty$

$x\to -\dfrac{\pi}{2}^{+}:\ \sec x+\tan x\to 0^{+} \Rightarrow g\to-\infty$

$g \text{ continuous increasing} \Rightarrow \text{range}=\mathbb{R} \Rightarrow (3) \text{ true}$

$\text{True: } (1),\ (2),\ (3);\qquad (4) \text{ is false}$