$\tan^{-1}(x^2-\frac{x^4}{2}+\frac{x^6}{4}-\cdots)+ \cot^{-1}(x^4-\frac{x^8}{2}+\frac{x^{12}}{4}-\cdots)=\pi/ 2$ 
Find $x$ with $0<|x|<2$ such that
  $$\tan^{-1}(x^2-\frac{x^4}{2}+\frac{x^6}{4}-\cdots)+ \cot^{-1}(x^4-\frac{x^8}{2}+\frac{x^{12}}{4}-\cdots)=\pi/ 2$$

My try:
$(x^2-\frac{x^4}{2}+\frac{x^6}{4}-\cdots)=\alpha$
$(x^4-\frac{x^8}{2}+\frac{x^{12}}{4}-\cdots)=\beta$
$\alpha=\frac{\pi }{2}-\beta \to \tan \alpha=\tan(\frac{\pi }{2}-\beta)=\cot \beta$
Now?

What I can come up with is as below
$\tan^{-1}(\alpha) =A \implies \tan(A)=\alpha$
$\cot^{-1}(\beta) =B \implies \cot(B)=\beta$
Also, we have
$A+B=\frac{\pi}{2}$
Add $\frac{\pi}{2}$ to both sides
$A+B+\frac{\pi}{2}=\pi$
Take $\tan$ from both sides and use the famous formula of $\tan(x+y)=\frac{\tan(x)+\tan(y)}{1-\tan(x)\tan(y)}$
$0=\tan(\pi)=\frac{\tan(A)+\tan(B+\frac{\pi}{2})}{1-\tan(A)\tan(B+\frac{\pi}{2})}$
Knowing $\tan(x+\frac{\pi}{2})=\cot(x)$, we get
$0=\tan(\pi)=\frac{\tan(A)+\cot(B)}{1-\tan(A)\cot(B)}=\frac{\alpha+\beta}{1-\alpha \beta}$
So, we need to have $\alpha+\beta=0$. Of course, we cannot have $1-\alpha \beta=0$, at the same time. Because, the n we have $\alpha=\frac{1}{\beta}$ and as we want to have $\alpha = -\beta$, by multiplying them we get the contradiction $\alpha^2=-1$.
Then, we need to find a closed formula for $(x^2-\frac{x^4}{2}+\frac{x^6}{4}-\cdots)=\alpha$ and $(x^4-\frac{x^8}{2}+\frac{x^{12}}{4}-\cdots)=\beta$, which can be found by separating each of them into positive and negative terms and using geometric series summation formula. Note that the limit on $0<|x|<2$ comes useful here, as we need the geometric summations to be convergent.
Finally, if I have not made any mistakes, you need to solve a polynomial of order $10$ to find $x$.
