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$\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.

N. F. Taussig
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Math-Learner
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1 Answers1

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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.

bjorn93
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