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How do you evaluate the above?

I know that $\cot(2\tan^{-1}(2)) = \cot(2\cot^{-1}\left(\frac{1}{2}\right))$, but I'm lost as to how to simplify this further.

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    $\begingroup$ Hint: $\tan^{-1}(a) +\tan^{-1}(b) = \tan^{-1}\left(\frac{a+b}{1-ab}\right)$ up to integer multiples of $\pi$. $\endgroup$ – achille hui May 26 '15 at 22:23
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Let $y=\cot(2\arctan 2)$. You can use the definition $\cot x\equiv \frac{1}{\tan x} $ and the identity for $\tan 2x\equiv\frac{2\tan x}{1-\tan^2 x}$ to find $\frac 1y=\tan(2 \arctan 2)$ in terms of $\tan(\arctan 2) = 2$.

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let $t = \arctan(2).$ then we have the following $$\tan(t) = 2, y=\sin t = 2/\sqrt 5, x=\cos t = 1/\sqrt 5, 0 < t < \pi/2. $$ to evaluate $$\cot(2t) = \frac{\cos (2t)}{\sin 2t} = \frac{x^2 - y^2}{2xy} = \frac{1-4}{2 \times 1 \times 2} = -\frac34. $$

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Use the formula $$\tan 2\alpha=\frac{2tan\alpha}{1-\tan^2\alpha}$$ $$\implies \cot(2\tan^{-1}(2))=\cot\left(\tan^{-1}\left(\frac{2(2)}{1-(2)^2}\right)\right)$$$$=\cot\left(\tan^{-1}\left(\frac{-4}{3}\right)\right)$$ $$=\cot\left(\pi-\tan^{-1}\left(\frac{4}{3}\right)\right)$$ $$=-\cot\left(\tan^{-1}\left(\frac{4}{3}\right)\right)$$ $$=-\cot\left(\cot^{-1}\left(\frac{3}{4}\right)\right)$$$$=-\frac{3}{4}=-0.75$$

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Why not simply evaluate the terms numerically, from inside out:

$ArcTan[2] = 1.10715$

$Cot[1.10715] = -0.75$.

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    $\begingroup$ Presumably, OP wants an analytical solution. $\endgroup$ – Brian Tung May 26 '15 at 22:42
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Let $\arctan 2=y\implies \tan y =2,\cot y=\dfrac12$

$\cot(2y)=\dfrac1{\tan2y}=\dfrac{1-\tan^2y}{2\tan y}=\dfrac{\cot^2y-1}{2\cot y}$

(multiplying the numerator & the denominator by $\cot^2y$)

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