$$f:\left( { - {\pi \over 2},{\pi \over 2}} \right) \to R$$

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

$$f( - x) = {[\log (\sec x - \tan x)]^3}$$

$$ = {\left[ {\log \left( {{{(\sec x - \tan x)(\sec x + \tan x)} \over {\sec x + \tan x}}} \right)} \right]^3}$$

$$ = {\left[ {\log \left( {{1 \over {\sec x + \tan x}}} \right)} \right]^3} = {[ - \log (\sec x + \tan x)]^3}$$

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

$$\therefore$$ f is an odd function. (a) is correct and (d) is not correct.

Also,

$$f'(x) = 3{[\log (\sec x + \tan x)]^2}\,.\,{{\sec x\tan x + {{\sec }^2}x} \over {\sec x + \tan x}}$$

$$ = 3\sec x{[\log (\sec x + \tan x)]^2} > 0\,\forall x \in \left( { - {\pi \over 2},{\pi \over 2}} \right)$$

$$\therefore$$ f is increasing on $$\left( { - {\pi \over 2},{\pi \over 2}} \right)$$

We know that strictly increasing function is one one.

$$\therefore$$ f is one one

$$\therefore$$ (b) is correct.

$$(\sec x + \tan x) = \tan \left( {{\pi \over 4} + {\pi \over 2}} \right)$$

as $$x \in \left( { - {\pi \over 2},{\pi \over 2}} \right)$$, then

$$0 < \tan \left( {{\pi \over 4} + {\pi \over 2}} \right) < \infty $$

$$0 < \sec x + \tan x < \infty $$

$$ \Rightarrow - \infty < \ln (\sec x + \tan x) < \infty $$

$$ - \infty < {[\ln (\sec x + \tan x)]^3} < \infty $$

$$ \Rightarrow - \infty < f(x) < \infty $$

Range of f(x) is R and thus f(x) is an onto function.

$$\therefore$$ (c) is correct.