# Why does \cot(2 \arctan (Ax))=\frac{1-(Ax)^2}{2Ax}

Why does $\mathrm{cot}\left(2\mathrm{arctan}\left(Ax\right)\right)=\frac{1-{\left(Ax\right)}^{2}}{2Ax}$
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vicki331g8
We have the following fundamental trigonometric identity:
$\mathrm{tan}2t=\frac{2\mathrm{tan}t}{1-{\mathrm{tan}}^{2}t}$
Take reciprocal on both sides:
$\frac{1}{\mathrm{tan}2t}=\frac{1-{\mathrm{tan}}^{2}t}{2\mathrm{tan}t}$
Now substitute $t=\mathrm{arctan}Ax$ to get the identity in the question. $\left(\mathrm{cot}x\equiv \frac{1}{\mathrm{tan}x}\right)$
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Jack Maxson
Use the fact that $\mathrm{cot}\theta =\frac{1}{\mathrm{tan}\theta }$
And the double angle formula for tangent:
$\mathrm{tan}2\theta =\frac{2\mathrm{tan}\theta }{1-{\mathrm{tan}}^{2}\theta }$
Lastly, before I confuse myself lets