Question

Trigonometría...........
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Answer 1

$\displaystyle\\E=\sum_{n=1}^4\arctan \dfrac{1}{1+n+n^2}\\ \\ \\E=\arctan \dfrac{1}{3}+\arctan \dfrac{1}{7}+\arctan \dfrac{1}{13}+\arctan \dfrac{1}{21}$

$E_1=\arctan\dfrac{1}{3}+\arctan\dfrac{1}{7}\\ \\ \\\tan E_1=\dfrac{1/3+1/7}{1-(1/3)(1/7)}=\dfrac{1}{2}\\ \\ \\E_2=\arctan\dfrac{1}{13}+\dfrac{1}{21}\\ \\ \\\tan E_2=\dfrac{1/13+1/21}{1-(1/13)(1/21)}=\dfrac{1}{8}$

Entonces

$E=\arctan\dfrac{1}{2}+\dfrac{1}{8}\\ \\ \\\tan E=\dfrac{1/2+1/8}{1-(1/2)(1/8)}=\dfrac{2}{3}\\ \\ \\E=\arctan\dfrac{2}{3}$