<s>[INST] If $\cot(A-B)=3$ and $\cot(A+B)=2$, then evaluate $\tan A + \frac{1}{\cot^2A}+\tan^3A+\frac{1}{\cot^4A}+\cdots$ 
Given:
$$\cot (A-B) = 3 \qquad \cot (A+B) = 2$$
Evaluate:
$$\tan A + \frac{1}{\cot^2 A} + \tan^3 A + \frac{1}{\cot^4 A} + \cdots $$

Attempts:
I think that the given information is used to get the value of $A$ so I can input $A$ to the problems and using the formula of infinite sum of geometric sequence. I haven't touched that step yet. I am still trying to get the value of $A$.
From
$$\cot (A-B) = \frac{\cot A \cot B-1}{\cot A + \cot B}$$
I get to $\cot A - 5\cot B = 2$.
Now, I am stuck. Where should I do next?
 [/INST] Let $\tan{A}=x$ and $\tan{B}=y$.
Thus, we obtain the following system.
$$\frac{x-y}{1+xy}=\frac{1}{3}$$ and
$$\frac{x+y}{1-xy}=\frac{1}{2}$$ or
$$3(x-y)=1+xy$$ and
$$2(x+y)=1-xy,$$ which gives
$$5x-y=2$$ or $$y=5x-2.$$
Now, since $$2(x+y)=1-xy,$$ we obtain:
$$2(x+5x-2)=1-x(5x-2)$$ or
$$12x-4=1-5x^2+2x$$ or
$$5x^2+10x-5$$ or
$$x^2+2x-1=0.$$
Can you end it now?
Actually.
For $\tan{A}=-1-\sqrt2$ your series diverges.  
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