cos^(−1) (((x^2 −1)/(x^2 +1)))+(1/2)tan^(−1) (((2x)/(1−x^2 )))=((2π)/3)


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Question Number 163936 by cortano1 last updated on 12/Jan/22

$\cos^{- 1} (\frac{x^{2} - 1}{x^{2} + 1}) + \frac{1}{2} \tan^{- 1} (\frac{2 x}{1 - x^{2}}) = \frac{2 π}{3}$
	Answered by mahdipoor last updated on 12/Jan/22

	$g e t c o s β = \frac{x^{2} - 1}{x^{2} + 1} \Rightarrow {t a n}^{2} β = \frac{1}{{c o s}^{2} β} - 1 = \frac{4 x^{2}}{{(x^{2} - 1)}^{2}}$ $\Rightarrow \pm t a n β = t a n \pm β = \frac{2 x}{1 - x^{2}}$ $i f 0 ⩽ β ⩽ \frac{π}{2} \Rightarrow$ $A = {c o s}^{- 1} (\frac{x^{2} - 1}{x^{2} + 1}) + 0.5 {t a n}^{- 1} (\frac{2 x}{1 - x^{2}}) =$ $β \pm 0.5 β = 1.5 π \Rightarrow β = π o r 3 π$ $i f π / 2 ⩽ β ⩽ π$ $A = β \pm 0.5 (β - π) = 1.5 π \Rightarrow β = \frac{4}{3} π o r 2 π$ $w i t h o u t a n s$
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