when x∈(−π2,π2), tan(x2+π4)∈(0,∞)
and from the above graph, clearly y=logx is one one for x∈(0,∞) as the graph is continuously increasing.
Hence, f(x) is a one-one function.
As for x∈(0,∞),logx∈(−∞,∞)
Similarly f(x)∈(−∞,∞)
Hence, f(x) is an onto function.
Now, f(x)=(log(secx+tanx))3 ⇒f(−x)=(log(secx−tanx))3 ⇒f(−x)=(log(1(secx+tanx)))3 f(−x)=−(log(secx+tanx))3=−f(x)
Hence, f(x) is an odd function.