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Question Number 132880 by faysal last updated on 17/Feb/21
cot(142.5°)=?
$${cot}\left(\mathrm{142}.\mathrm{5}°\right)=? \\ $$
Commented by bramlexs22 last updated on 17/Feb/21
285°= 2×142.5 285°=270°+15° cot 285° = cot 285° cot (270°+15°)=cot (2×142.5°) −tan 15°= cot 2α ; α=142.5° ⇔cot 2α=−(((1−(1/( (√3))))/(1+(1/( (√3))))))=−(((√3)−1)/( (√3)+1)) ⇔tan 2α = (((√3)+1)/(1−(√3))) = ((4+2(√3))/(−2)) ((2tan α)/(1−tan^2 α)) = −(√3)−2 ⇒2tan α = (2+(√3))tan^2 α−((√3)+2) (2+(√3))tan^2 α−2tan α−((√3)+2)=0 tan α=((2−(√(4+4(2+(√3))^2 )))/(2(2+(√3)))) tan α= ((2−2(√(8+4(√3))))/(2(2+(√3))))=((1−2(√(2+(√3))))/(2+(√3))) cot α= ((2+(√3))/(1−2(√(2+(√3))))) . so cot 142.5° = ((2+(√3))/(1−2(√(2+(√3))))) ≈−1.303225
$$\mathrm{285}°=\:\mathrm{2}×\mathrm{142}.\mathrm{5} \\ $$$$\mathrm{285}°=\mathrm{270}°+\mathrm{15}° \\ $$$$\mathrm{cot}\:\mathrm{285}°\:=\:\mathrm{cot}\:\mathrm{285}° \\ $$$$\mathrm{cot}\:\left(\mathrm{270}°+\mathrm{15}°\right)=\mathrm{cot}\:\left(\mathrm{2}×\mathrm{142}.\mathrm{5}°\right) \\ $$$$−\mathrm{tan}\:\mathrm{15}°=\:\mathrm{cot}\:\mathrm{2}\alpha\:;\:\alpha=\mathrm{142}.\mathrm{5}° \\ $$$$\Leftrightarrow\mathrm{cot}\:\mathrm{2}\alpha=−\left(\frac{\mathrm{1}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}}{\mathrm{1}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}}\right)=−\frac{\sqrt{\mathrm{3}}−\mathrm{1}}{\:\sqrt{\mathrm{3}}+\mathrm{1}} \\ $$$$\Leftrightarrow\mathrm{tan}\:\mathrm{2}\alpha\:=\:\frac{\sqrt{\mathrm{3}}+\mathrm{1}}{\mathrm{1}−\sqrt{\mathrm{3}}}\:=\:\frac{\mathrm{4}+\mathrm{2}\sqrt{\mathrm{3}}}{−\mathrm{2}} \\ $$$$\:\frac{\mathrm{2tan}\:\alpha}{\mathrm{1}−\mathrm{tan}\:^{\mathrm{2}} \alpha}\:=\:−\sqrt{\mathrm{3}}−\mathrm{2} \\ $$$$\Rightarrow\mathrm{2tan}\:\alpha\:=\:\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)\mathrm{tan}\:^{\mathrm{2}} \alpha−\left(\sqrt{\mathrm{3}}+\mathrm{2}\right) \\ $$$$\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)\mathrm{tan}\:^{\mathrm{2}} \alpha−\mathrm{2tan}\:\alpha−\left(\sqrt{\mathrm{3}}+\mathrm{2}\right)=\mathrm{0} \\ $$$$\mathrm{tan}\:\alpha=\frac{\mathrm{2}−\sqrt{\mathrm{4}+\mathrm{4}\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)^{\mathrm{2}} }}{\mathrm{2}\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)} \\ $$$$\mathrm{tan}\:\alpha=\:\frac{\mathrm{2}−\mathrm{2}\sqrt{\mathrm{8}+\mathrm{4}\sqrt{\mathrm{3}}}}{\mathrm{2}\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)}=\frac{\mathrm{1}−\mathrm{2}\sqrt{\mathrm{2}+\sqrt{\mathrm{3}}}}{\mathrm{2}+\sqrt{\mathrm{3}}} \\ $$$$\mathrm{cot}\:\:\alpha=\:\frac{\mathrm{2}+\sqrt{\mathrm{3}}}{\mathrm{1}−\mathrm{2}\sqrt{\mathrm{2}+\sqrt{\mathrm{3}}}}\:. \\ $$$$\mathrm{so}\:\mathrm{cot}\:\mathrm{142}.\mathrm{5}°\:=\:\frac{\mathrm{2}+\sqrt{\mathrm{3}}}{\mathrm{1}−\mathrm{2}\sqrt{\mathrm{2}+\sqrt{\mathrm{3}}}}\:\approx−\mathrm{1}.\mathrm{303225} \\ $$$$ \\ $$
Commented by bramlexs22 last updated on 17/Feb/21

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