Let the function \(\mathrm{g}:(-\infty, \infty) \rightarrow\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\) be given by \(\mathrm{g}(\mathrm{u})=2 \tan ^{-1}\left(\mathrm{e}^{\mathrm{u}}\right)-\frac{\pi}{2}\). Then g is

\(\begin{aligned}
& \mathrm{g}(\mathrm{u})=2 \tan ^{-1}\left(\mathrm{e}^{\mathrm{u}}\right)-\frac{\pi}{2} \\
& \Rightarrow \mathrm{~g}(\mathrm{u})=\tan ^{-1}\left(\mathrm{e}^{\mathrm{u}}\right)-\left(\frac{\pi}{2}-\tan ^{-1}\left(\mathrm{e}^{\mathrm{u}}\right)\right) \\
& \Rightarrow \mathrm{g}(\mathrm{u})=\tan ^{-1}\left(\mathrm{e}^{\mathrm{u}}\right)-\cot ^{-1}\left(e^{\mathrm{u}}\right) \\
& \Rightarrow \mathrm{g}(-\mathrm{u})=\tan ^{-1}\left(\mathrm{e}^{-\mathrm{u}}\right)-\cot ^{-1}\left(\mathrm{e}^{-\mathrm{u}}\right) \\
& \Rightarrow \mathrm{g}(-\mathrm{u})=\tan ^{-1}\left(\frac{1}{\mathrm{e}^{\mathrm{u}}}\right)-\cot ^{-1}\left(\frac{1}{\mathrm{e}^u}\right) \\
& \Rightarrow \mathrm{g}(-\mathrm{u})=\cot ^{-1}\left(\mathrm{e}^{\mathrm{u}}\right)-\tan ^{-1}\left(\mathrm{e}^{\mathrm{u}}\right)=-\mathrm{g}(\mathrm{u})
\end{aligned}\)
Now, \(\dot{g(\mathrm{u}})=2 \tan ^{-1}\left(\mathrm{e}^{\mathrm{u}}\right)-\frac{\pi}{2}\)
\(\Rightarrow \mathrm{g}^{\prime}(\mathrm{u})=\frac{2 \mathrm{e}^{\mathrm{u}}}{1+\mathrm{e}^{2 \mathrm{u}}}\gt0 \text { for all } \mathrm{u} \in(-\infty, \infty)\)
\(\Rightarrow \mathrm{g}\) is strictly increasing function in \((-\infty, \infty)\). Hence, \(\mathrm{g}(\mathrm{u})\) is odd and is strictly increasing in \((-\infty, \infty)\).