Let f: (- (π)/(2), (π)/(2)) → R be given by f(x) = ( (sec x + tan...

f\left(x\right) = {\left(\log \left(sec x + tan x\right)\right)}^{3} ∀ x ∈ \left(- \dfrac{π}{2}, \dfrac{π}{2}\right)$<br />$f\left(- x\right) = - f\left(x\right),$hence$f\left(x\right)$is odd function<br />Let$g\left(x\right) = sec x + tan x ∀ x ∈ \left(- \dfrac{π}{2}, \dfrac{π}{2}\right)$<br />$\to {g}^{′}\left(x\right) = sec x\left(sec x + tan x\right) > 0 ∀ x ∈ \left(- \dfrac{π}{2}, \dfrac{π}{2}\right)$<br />$\to g\left(x\right)$is one - one function<br />Hence${\left(\log_{e} \left(g\left(x\right)\right)\right)}^{3} is one - one function.
And g\left(x\right) &isin; \left(0, \infty\right) &forall; x &isin; \left(- \dfrac{&pi;}{2}, \dfrac{&pi;}{2}\right)
$\to$ \log \left(g\left(x\right)\right) &isin; R Hence $f\left(x\right)$ is an onto function.