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

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

$$\:\mathrm{cos}^{−\mathrm{1}} \left(\frac{{x}^{\mathrm{2}} −\mathrm{1}}{{x}^{\mathrm{2}} +\mathrm{1}}\right)+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{2}{x}}{\mathrm{1}−{x}^{\mathrm{2}} }\right)=\frac{\mathrm{2}\pi}{\mathrm{3}} \\ $$

Answered by mahdipoor last updated on 12/Jan/22

get cosβ=((x^2 −1)/(x^2 +1)) ⇒ tan^2 β=(1/(cos^2 β))−1=((4x^2 )/((x^2 −1)^2 )) ⇒±tanβ=tan±β=((2x)/(1−x^2 )) if 0≤β≤(π/2) ⇒ A=cos^(−1) (((x^2 −1)/(x^2 +1)))+0.5tan^(−1) (((2x)/(1−x^2 )))= β±0.5β=1.5π ⇒β=π or 3π if π/2≤β≤π A=β±0.5(β−π)=1.5π ⇒ β=(4/3)π or 2π without ans

$${get}\:{cos}\beta=\frac{{x}^{\mathrm{2}} −\mathrm{1}}{{x}^{\mathrm{2}} +\mathrm{1}}\:\:\Rightarrow\:{tan}^{\mathrm{2}} \beta=\frac{\mathrm{1}}{{cos}^{\mathrm{2}} \beta}−\mathrm{1}=\frac{\mathrm{4}{x}^{\mathrm{2}} }{\left({x}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\Rightarrow\pm{tan}\beta={tan}\pm\beta=\frac{\mathrm{2}{x}}{\mathrm{1}−{x}^{\mathrm{2}} } \\ $$$${if}\:\:\mathrm{0}\leqslant\beta\leqslant\frac{\pi}{\mathrm{2}}\:\Rightarrow \\ $$$${A}={cos}^{−\mathrm{1}} \left(\frac{{x}^{\mathrm{2}} −\mathrm{1}}{{x}^{\mathrm{2}} +\mathrm{1}}\right)+\mathrm{0}.\mathrm{5}{tan}^{−\mathrm{1}} \left(\frac{\mathrm{2}{x}}{\mathrm{1}−{x}^{\mathrm{2}} }\right)= \\ $$$$\beta\pm\mathrm{0}.\mathrm{5}\beta=\mathrm{1}.\mathrm{5}\pi\:\Rightarrow\beta=\pi\:{or}\:\mathrm{3}\pi\: \\ $$$${if}\:\:\pi/\mathrm{2}\leqslant\beta\leqslant\pi\: \\ $$$${A}=\beta\pm\mathrm{0}.\mathrm{5}\left(\beta−\pi\right)=\mathrm{1}.\mathrm{5}\pi\:\Rightarrow\:\beta=\frac{\mathrm{4}}{\mathrm{3}}\pi\:{or}\:\mathrm{2}\pi \\ $$$${without}\:{ans}\:\: \\ $$

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