**Correct option (a,b,c)**

**Explanation :**

(-π/2,π/2) → R

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

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But at x = 0,

log (sec x + tan x) = ∞

⇒ {log(secx + tanx)}^{2} > 0

Therefore, f(x) is strictly increasing in (-π/2,π/2).

Now when x → π^{-}/2,f(x) → ∞ and π → π^{+}/2,f(x) → -∞

Therefore, f(x) being continuous in its domain, it covers whole co-domain, i.e. R. See Fig.Hence, it is onto

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