How to prove a hard inverse trigonometry mathematical induction $\arctan\left(\frac{1}{2\left(1\right)^{2}}\right)+\arctan\left(\frac{1}{2\left(2\right)^{2}}\right)+\arctan\left(\frac{1}{2\left(3\right)^{2}}\right)+...+\arctan\left(\frac{1}{2\left(n\right)^{2}}\right)=\frac{\pi}{4}-\arctan\left(\frac{1}{2n+1}\right)$
I assume you need to do something like "take tan of both sides" but that doesn't really work for me given the summation on the LHS. Any hints?

I'll assume you were able to prove the base case $n = 1$.
The inductive hypothesis is to assume $$\arctan \dfrac{1}{2\cdot1^2}+\cdots+\arctan \dfrac{1}{2\cdot n^2} = \dfrac{\pi}{4} - \arctan \dfrac{1}{2n+1}. \quad (1)$$
Then, you need to prove $$\arctan \dfrac{1}{2\cdot1^2}+\cdots+\arctan \dfrac{1}{2\cdot n^2} +\arctan \dfrac{1}{2\cdot (n+1)^2} = \dfrac{\pi}{4} - \arctan \dfrac{1}{2n+3}. \quad (2)$$
If you subtract $(1)$ from $(2)$, you get the equation $$\arctan \dfrac{1}{2\cdot (n+1)^2} = \arctan \dfrac{1}{2n+1} - \arctan \dfrac{1}{2n+3}. \quad (3)$$
So, after assuming the inductive hypothesis $(1)$, you need to prove equation $(3)$. Then you can add equation $(3)$ to equation $(1)$ to prove equation $(2)$, which completes the inductive step.
