Prove that 16arctan((1/5))−4arctan((1/(239)))=π


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Question Number 79124 by ~blr237~ last updated on 22/Jan/20

$P r o v e t h a t$ $16 a r c t a n (\frac{1}{5}) - 4 a r c t a n (\frac{1}{239}) = π$
Commented by mind is power last updated on 23/Jan/20

$n i c e o n e$
	Answered by ~blr237~ last updated on 27/Jan/20

	$l e t t a k e z = a + i b w i t h a \neq 0$ $w e c a n p r o v e t h a t a r z \equiv a r c t a n (\frac{b}{a}) [2 π]$ $l e t n a m e d A = 4 a r c t a n (\frac{1}{5}) - a r c t a n (\frac{1}{239})$ $A \equiv 4 a r g (5 + i) - a r g (239 + i) [2 π]$ $A \equiv a r g [{(5 + i)}^{4}] - a r g [(239 + i)] [2 π]$ $A \equiv a r g [((476 + 480 i)] - a r g (239 + i) [2 π]$ $A \equiv a r g [\frac{(476 + 480 i) (239 - i)}{239^{2} - 1}] [2 π]$ $A \equiv a r g [\frac{476 \times 239 + 480 + i (480 \times 239 - 476)}{238 \times 240}] [2 π]$ $A \equiv a r g [\frac{114244 + 114244 i}{57120}] [2 π]$ $A \equiv a r g [\frac{114244}{57120} (1 + i)] [2 π]$ $A \equiv 0 + a r g (1 + i) [2 π]$ $A \equiv \frac{π}{4} [2 π]$ $s u c h a s a r c t a n (\frac{1}{239}), a r c t a n (\frac{1}{5}) \in [0, \frac{π}{2}] w e h a v e a f t e r f r a m i n g A : - \frac{π}{2} ⩽ A ⩽ 2 π$ $S o f i n a l y A = \frac{π}{4}$
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