Calculator is for the dull - 4

Generally ,suppose $\tan^{-1}(x)-\phi=φ$

$\implies x=\tan(\phi+φ) \implies x=\dfrac{\tan(\phi)+\tan(φ)}{1-\tan(\phi)\tan(φ)}$

Here $4\tan^{-1}(\frac{3}{2}) = 2\tan^{-1}(\frac{3}{2})+2\tan^{-1}(\frac{3}{2})$

$=\tan^{-1}(\frac{3}{1-\frac{9}{4}} )+\tan^{-1}(\frac{3}{1-\frac{9}{4}})$

$=-\tan^{-1}(\frac{\frac{24}{5}}{1-\frac{144}{25}}) =\tan^{-1}(\frac{120}{119})$

Now $x=\dfrac{120}{119}$ And $\phi=\dfrac{5π}{4}$

So $\dfrac{120}{119}= \dfrac{1+\tan(φ)}{1-\tan(φ)} \implies \dfrac{120+119}{120-119}= \cot(φ)=239$

Αctual question was about finding $\cot(φ)$ . So answer is $\boxed{239}$

$A = 4 \arctan{1.5} - \frac{5\pi}{4}$ $A = 4 \arctan{1.5} - 4 \arctan{1} - \arctan{1}$ $A = 4 \arctan\left(\frac{1.5 - 1}{1 + 1.5}\right) - \arctan{1}$ $A = 4 \arctan\left(\frac{1}{5}\right) - \arctan{1}$ $A = 2\left(\arctan\left(\frac{1}{5}\right) + \arctan\left(\frac{1}{5}\right)\right) - \arctan{1}$ $A = 2 \arctan\left(\frac{2/5}{1 - 1/25}\right)- \arctan{1}$ $A = 2 \arctan\left(\frac{5}{12}\right) - \arctan{1}$ $A = \arctan\left(\frac{5}{12}\right) + \arctan\left(\frac{5}{12}\right) - \arctan{1}$ $A = \arctan\left(\frac{10/12}{1 - 25/144}\right) - \arctan{1}$ $A = \arctan\left(\frac{120}{119}\right)- \arctan{1}$ $A = \arctan\left(\frac{120/119 -1}{1 + 120/119}\right)$ $A = \arctan\left(\frac{1}{239}\right)$ $\tan{A} = \frac{1}{239}$ $\implies \boxed{\cot{A} = 239}$

$\begin{aligned} \cot \left(4 \tan^{-1} \frac 32 - \frac {5\pi}4 \right) & = \cot \left(4 \theta - \frac {5\pi}4 \right) & \small \blue{\text{Let }\tan \theta = \frac 32} \\ & = \tan \left(\frac {7\pi}4 - 4 \theta \right) & \small \blue{\text{Note that } \cot \phi = \tan \left(\frac \pi 2 - \phi \right)} \\ & = \tan \left(-\frac \pi 4 -4 \theta \right) \\ & = -\tan \left(\frac \pi 4 + 4 \theta \right) \\ & = \frac {\tan 4\theta+1}{\tan 4\theta-1} \\ & = \frac {\frac {2\tan 2\theta}{1-\tan^2 2\theta}+1}{\frac {2\tan 2\theta}{1-\tan^2 2\theta}-1} & \small \blue{\text{also }\tan 2\theta = \frac {2 \cdot \frac 32}{1-\frac 94} = - \frac {12}5} \\ & = \frac {\frac {120}{119}+1}{\frac {120}{119}-1} \\ & = \frac {120+ 119}{120 - 119} \\ & = \boxed{239} \end{aligned}$