Finding whether the inverse trigonometry function is odd or even

Question

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

(A) even and strictly increasing in $(0,\infty)$

(B) odd and strictly decreasing in $(-\infty,\infty)$

(C) odd and strictly increasing in $(-\infty,\infty)$

(D) neither even nor odd and is strictly increasing in $(-\infty,\infty)$

The official answer key is (C) odd and strictly increasing in $(-\infty,\infty)$

My approach is as follow

$g\left( u \right) = 2{\tan ^{ - 1}}{e^u} - \frac{\pi }{2} = {\tan ^{ - 1}}\frac{{2{e^u}}}{{1 - {e^{2u}}}} - \frac{\pi }{2}$

$g'\left( u \right) = \frac{2}{{1 + {e^{2u}}}}$, hence it is an increasing function when $u \in R$

$g\left( { - u} \right) = 2{\tan ^{ - 1}}{e^{ - u}} - \frac{\pi }{2} = {\tan ^{ - 1}}\frac{{2{e^{ - u}}}}{{1 - {e^{ - 2u}}}} - \frac{\pi }{2} = {\tan ^{ - 1}}\frac{{2{e^{ - u}}{e^{2u}}}}{{{e^{2u}} - 1}} - \frac{\pi }{2} = {\tan ^{ - 1}}\frac{{ - 2{e^u}}}{{1 - {e^{2u}}}} - \frac{\pi }{2}$

$g\left( { - u} \right) = - {\tan ^{ - 1}}\frac{{2{e^u}}}{{1 - {e^{2u}}}} + \frac{\pi }{2} - \pi = - g\left( u \right) - \pi $

For odd function $g(u)=-g(-u)$

I consulted a book to see that the following is valid $\tan^{-1}(-x)=-\tan^{-1}x$ for $x \in R$ and $\cot^{-1}(-x)=\pi-\cot^{-1}x$ for $x \in\Bbb R$.u)

I am getting $g(-u)=-g(u)+\pi$

Answer 1

We have

$\tan^{-1}(\frac1x) = \begin{cases}\cot^{-1}(x) & \text{if }x>\>0 \\\cot^{-1}(x) - \pi &\text{if }x<0\end{cases}$

So, $\color{blue}{2\tan^{-1}(e^{-u})} =2\tan^{-1}(\frac1{e^u}) = 2(\cot^{-1}(e^u)) = 2\left[\frac\pi2 - \tan^{-1}(e^u)\right] = \color{blue}{\pi - 2\tan^{-1}(e^u)}$.

Now,

$g(-u) =2\tan^{-1}(e^{-u}) -\frac\pi2 = \pi - 2\tan^{-1}(e^u) -\frac\pi2= \color{blue}{\frac\pi2-2\tan^{-1}(e^u)} = -g(u)$.

Answer 2

For $g(u)$ to be odd we should have

$$ 2{\tan ^{ - 1}}{e^u} - \frac{\pi }{2}= -2[{\tan ^{ - 1}}{e^{-u}} - \frac{\pi }{2} ]$$

i.e., (transpose, divide by 2) we should have

$${\tan ^{ - 1}{e^u}+\tan ^{ - 1}{e^{-u}}} = \pi/2 $$

Taking tan of either side we should have

$$\dfrac{e^u+e^{-u}}{1-1} = \infty $$

and as we have it so indeed, and along with your showing it to be monotonous, the given proposition is correct.