Show that $\arctan\left({1\over A}\right)=\arctan\left({1\over B}\right)+\arctan\left({1\over C}\right)$ $1,1,2,3,5,8,...$ for $n=1,2,3,4,5,...$ ;It is the n-th Fibonacci numbers.
Show that,
$\arctan\left({1\over xF_{2n-1}+F_{2n}}\right)=\arctan\left({1\over xF_{2n-1}+F_{2n+1}}\right)+\arctan\left({1\over x^2F_{2n-1}+xF_{2n+2}+F_{2n+2}}\right)$
I try:
Using 
$$\arctan\left({1\over a}\right)-\arctan\left({1\over b}\right)=\arctan\left({b-a\over ab+1}\right)$$
Just ignore the arctan for the sake of space.
$$\arctan{1\over xF_{2n-1}+F_{2n}}-\arctan{1\over xF_{2n-1}+F_{2n+1}}=\arctan{1\over x^2F_{2n-1}+xF_{2n+2}+F_{2n+2}}=T$$
$$\arctan{F_{2n+1}-F_{2n}\over x^2F_{2n-1}^2+xF_{2n-1}(F_{2n}+F_{2n+1})+F_{2n}F_{2n+1}+1}=T$$
$$\arctan{F_{2n+1}-F_{2n}\over x^2F_{2n-1}^2+xF_{2n-1}F_{2n+2}+F_{2n}F_{2n+1}+1}=T$$
$$\arctan{F_{2n-1}\over x^2F_{2n-1}^2+xF_{2n-1}F_{2n+2}+F_{2n}F_{2n+1}+1}=T$$
Any hints on how to simplify this further?

We make use of the identity:$$F_{2n-1}^2 -F_{2n}^2 + F_{2n}F_{2n-1} = 1...(1)$$ We can rewrite this as: $$2F_{2n}F_{2n-1} + F_{2n-1}^2 = F_{2n}^2 + F_{2n}F_{2n-1} + 1$$ $$\implies F_{2n-1}[2F_{2n} + F_{2n-1}] = F_{2n}[F_{2n} + F_{2n-1}] + 1...(2)$$ Now, we use the identity: $$F_{n+m} = F_nF_{m+1} + F_{n-1}F_m$$ to change the bracketed terms of $(2)$ into: $$F_{2n-1}F_{2n+2} = F_{2n}F_{2n+1} + 1$$ This is the same condition that is obtained by equating the $\arctan \frac{F_{2n-1}}{x^2F_{2n-1}^2 + xF_{2n-1}F_{2n+2} + F_{2n}F_{2n+1} + 1}$ and $\arctan \frac{1}{x^2F_{2n-1} + xF_{2n+2} +F_{2n+2}}$.   
Hope it helps.
