This question is mostly out of curiosity, but I hope that the answer will advance my understanding of the trigonometric function. While playing around with trigonometric functions on WolframAlpha I stumbled upon this $$\cot(2\arctan(Ax))=\frac{1-(Ax)^2}{2Ax}$$ I suspect that one could prove this using the complex forms of the trigonometric functions $\sin(z) = \dfrac{e^{iz}-e^{-iz}}{2i}$ but I myself do not know the complex form of $\arctan$ so I am unable to do this.

Is there some nice geometric or algebraic proof that explains why $\cot(2\arctan(Ax))=\frac{1-(Ax)^2}{2Ax}$? Or maybe, is WolframAlpha wrong and does $\cot(2\arctan(Ax))$ just happen to be very close to $\frac{1-(Ax)^2}{2Ax}$?


2 Answers 2


We have the following fundamental trigonometric identity: $$\tan2t=\frac{2\tan t}{1-\tan^2t}$$ Take reciprocal on both sides: $$\frac1{\tan2t}=\frac{1-\tan^2t}{2\tan t}$$ Now substitute $t=\arctan Ax$ to get the identity in the question. ($\cot x\equiv\frac1{\tan x}$.)


Use the fact that $\cot \theta=\dfrac{1}{\tan \theta}$

And the double angle formula for tangent:

$\tan 2\theta=\dfrac{2\tan \theta}{1-\tan^2 \theta}$

Lastly, before I confuse myself let's do some replacement.

$\tan^{-1} Ax=z$

$Ax=\tan z$

Hence the full thing is:

$=\left(\tan (2\tan^{-1} Ax) \right)^{-1}$

$=(\tan 2z)^{-1}= \dfrac{1-\tan^2 z}{2\tan z}$

All good now, I hope.

Substitute $\tan z$ for $Ax$


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