If $\cot(A-B)=3$ and $\cot(A+B)=2$, then evaluate $\tan A + \frac{1}{\cot^2A}+\tan^3A+\frac{1}{\cot^4A}+\cdots$

Question

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?

Answer 1

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.

Answer 2

Hint:

There must be something wrong in the question

$\tan2A=\tan(A+B+A-B)=\cdots=1$

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

One of the roots if $a=\sqrt2-1,|a|=\dfrac1{\sqrt2+1}<1$

The other root is $b=-\dfrac1a$

$|b|=\dfrac1{|a|}>1$

If $\tan A=p$

$$\sum_{m=0}^\infty p^{2m+1}=\dfrac p{1-p^2}$$ provided $p^2<1$

$$\sum_{n=0}\dfrac1{p^{2n}}=\dfrac{1/p^2}{1-1/p^2}$$ provided $\dfrac1{p^2}<1\iff p^2>1$

So, both infinite series cannot converge simultaneously