$\tan ^{-1}(\frac 12 \tan 2A)+\tan^{-1} ( \cot A) +\tan ^{-1}(\cot ^3 A)=\begin{cases} 0 & \frac {\pi} 4\lt A \le \frac {\pi} 2\\ \pi & 0\le A \le \frac {\pi} 2 \end {cases} $
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$\tan ^{-1}(\cot A) +\tan^{-1} (\cot^3 A) =\tan^{-1} (-\frac 12 \tan 2A) $
Now the problem reduces to finding the value of
$\tan ^{-1}(\frac 12 \tan 2A)+\tan ^{-1}(-\frac 12 \tan 2A)$
Now when $\frac {\pi} 4\lt A \le \frac {\pi} 2$ then $2A$ lies in the second quadrant, so the value of $\tan 2A $ is negative there, hence the value of $\frac 12 \tan 2A$ will be negative and the value of $-\frac 12 \tan 2A$ will be positive. I'm stucked here. How to justify the next steps from this calculation in the way I'm trying to justify by considering quadrant.
The identity for adding inverse tangents has cases: $$\tan^{-1}x+\tan^{-1}y= \begin{cases} \tan^{-1}\frac{x+y}{1-xy} & xy < 1 \\ \pi+\tan^{-1}\frac{x+y}{1-xy} & xy > 1,x>0 \\ -\pi+\tan^{-1}\frac{x+y}{1-xy} & xy > 1,x<0 \\ \end{cases}$$ That gives us $\tan^{-1}(\cot A)+\tan^{-1}(\cot^3 A)=\varepsilon\pi+\tan^{-1}\left(-\frac 12\tan 2A\right)$ where $\varepsilon=1$ if $A\in (0,\pi/4)$, $\varepsilon=0$ if $A\in(\pi/4,3\pi/4)$ and $\varepsilon=-1$ if $A\in(3\pi/4,\pi)$. For example, when $A\in(0,\pi/4)$, we have $\cot A>1\Rightarrow \cot^3 A>1$ so $\cot A\cdot\cot^3 A>1$ and we end up in the second case of the identity above. Since the inverse tangent is an odd function, i.e. $\tan^{-1}(-t)=-\tan^{-1}t$, adding $\tan^{-1}\left(\frac 12\tan 2A\right)$ cancels out $\tan^{-1}\left(-\frac 12\tan 2A\right)$ so the final result is: $$\tan^{-1}\left(\frac 12\tan 2A\right)+\tan^{-1}(\cot A)+\tan^{-1}(\cot^3 A)= \begin{cases} \pi, & A\in\left(0,\frac{\pi}{4}\right) \\ 0, & A\in\left(\frac{\pi}{4},\frac{3\pi}{4}\right) \\ -\pi, & A\in\left(\frac{3\pi}{4},\pi\right) \end{cases}$$ Notice that the expression is $\pi$-periodic so it's sufficient to evaluate it in those three intervals.