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Question Number 109914 by 1549442205PVT last updated on 26/Aug/20

Prove that tan142°30′+(√6)+(√3)−(√2) is an integer.

$$\mathrm{Prove}\:\mathrm{that}\:\mathrm{tan142}°\mathrm{30}'+\sqrt{\mathrm{6}}+\sqrt{\mathrm{3}}−\sqrt{\mathrm{2}} \\ $$$$\mathrm{is}\:\mathrm{an}\:\mathrm{integer}. \\ $$

Answered by Dwaipayan Shikari last updated on 26/Aug/20

−tan(37.5)°=−tan((5π)/(24)) 1−cos((5π)/(12))=2sin^2 ((5π)/(24)) tan((5π)/(24))=((sin((5π)/(24)))/(cos((5π)/(24))))=((2sin^2 ((5π)/(24)))/(2cos((5π)/(24))sin((5π)/(24))))=((1−cos((5π)/(12)))/(sin((5π)/(12))))=((1−(((√3)−1)/(2(√2))))/( (((√3)+1)/(2(√2)))))=((2(√2)−(√3)+1)/( (√3)+1)) =((2(√2)−(√3)+1)/( 2))((√3)−1)=((2(√6)−2(√2)−(4−2(√3)))/2)=(√6)−(√2)−2+(√3) −tan((5π)/(24))=−(√6)+(√2)+2−(√3) tan142.5°+(√6)+(√3)−(√2)=2

$$−{tan}\left(\mathrm{37}.\mathrm{5}\right)°=−{tan}\frac{\mathrm{5}\pi}{\mathrm{24}} \\ $$$$\mathrm{1}−{cos}\frac{\mathrm{5}\pi}{\mathrm{12}}=\mathrm{2}{sin}^{\mathrm{2}} \frac{\mathrm{5}\pi}{\mathrm{24}} \\ $$$${tan}\frac{\mathrm{5}\pi}{\mathrm{24}}=\frac{{sin}\frac{\mathrm{5}\pi}{\mathrm{24}}}{{cos}\frac{\mathrm{5}\pi}{\mathrm{24}}}=\frac{\mathrm{2}{sin}^{\mathrm{2}} \frac{\mathrm{5}\pi}{\mathrm{24}}}{\mathrm{2}{cos}\frac{\mathrm{5}\pi}{\mathrm{24}}{sin}\frac{\mathrm{5}\pi}{\mathrm{24}}}=\frac{\mathrm{1}−{cos}\frac{\mathrm{5}\pi}{\mathrm{12}}}{{sin}\frac{\mathrm{5}\pi}{\mathrm{12}}}=\frac{\mathrm{1}−\frac{\sqrt{\mathrm{3}}−\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}}{\:\frac{\sqrt{\mathrm{3}}+\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}}=\frac{\mathrm{2}\sqrt{\mathrm{2}}−\sqrt{\mathrm{3}}+\mathrm{1}}{\:\sqrt{\mathrm{3}}+\mathrm{1}} \\ $$$$=\frac{\mathrm{2}\sqrt{\mathrm{2}}−\sqrt{\mathrm{3}}+\mathrm{1}}{\:\mathrm{2}}\left(\sqrt{\mathrm{3}}−\mathrm{1}\right)=\frac{\mathrm{2}\sqrt{\mathrm{6}}−\mathrm{2}\sqrt{\mathrm{2}}−\left(\mathrm{4}−\mathrm{2}\sqrt{\mathrm{3}}\right)}{\mathrm{2}}=\sqrt{\mathrm{6}}−\sqrt{\mathrm{2}}−\mathrm{2}+\sqrt{\mathrm{3}} \\ $$$$−{tan}\frac{\mathrm{5}\pi}{\mathrm{24}}=−\sqrt{\mathrm{6}}+\sqrt{\mathrm{2}}+\mathrm{2}−\sqrt{\mathrm{3}} \\ $$$${tan}\mathrm{142}.\mathrm{5}°+\sqrt{\mathrm{6}}+\sqrt{\mathrm{3}}−\sqrt{\mathrm{2}}=\mathrm{2} \\ $$

Commented by 1549442205PVT last updated on 27/Aug/20

Thank Sir.

$$\mathrm{Thank}\:\mathrm{Sir}. \\ $$

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