<s>[INST] 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?
 [/INST] 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.
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