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Question Number 63363 by rajesh4661kumar@gamil.com last updated on 03/Jul/19

Commented by Prithwish sen last updated on 03/Jul/19

tan70=((tan40+tan20+tan10−tan40tan20tan10)/(1−tan40tan20−tan40tan10−tan20tan10)) tan70 − tan70tan40tan20−tan70tan40tan10−tan70tan20tan10 = tan40+tan20+tan10−tan40tan20tan10 ∵tan70tan20=1 ∴tan70tan40tan20=tan40 and tan70tan20tan10 = tan10 ⇒tan70−2tan40−tan20=2tan10+tan40tan10(tan70−tan20) =2tan10+tan40tam10(((1−tan^2 20)/(tan20)) )=2tan10+tan10(((2tan20)/(1−tan^2 20))) (((1−tan^2 20)/(tan20)))=4tan10 proved.

$$\mathrm{tan70}=\frac{\mathrm{tan40}+\mathrm{tan20}+\mathrm{tan10}−\mathrm{tan40tan20tan10}}{\mathrm{1}−\mathrm{tan40tan20}−\mathrm{tan40tan10}−\mathrm{tan20tan10}} \\ $$$$\mathrm{tan70}\:−\:\mathrm{tan70tan40tan20}−\mathrm{tan70tan40tan10}−\mathrm{tan70tan20tan10} \\ $$$$=\:\mathrm{tan40}+\mathrm{tan20}+\mathrm{tan10}−\mathrm{tan40tan20tan10} \\ $$$$\because\mathrm{tan70tan20}=\mathrm{1}\:\therefore\mathrm{tan70tan40tan20}=\mathrm{tan40}\:\mathrm{and}\:\mathrm{tan70tan20tan10}\:=\:\mathrm{tan10} \\ $$$$\Rightarrow\mathrm{tan70}−\mathrm{2tan40}−\mathrm{tan20}=\mathrm{2tan10}+\mathrm{tan40tan10}\left(\mathrm{tan70}−\mathrm{tan20}\right) \\ $$$$=\mathrm{2tan10}+\mathrm{tan40tam10}\left(\frac{\mathrm{1}−\mathrm{tan}^{\mathrm{2}} \mathrm{20}}{\mathrm{tan20}}\:\right)=\mathrm{2tan10}+\mathrm{tan10}\left(\frac{\mathrm{2tan20}}{\mathrm{1}−\mathrm{tan}^{\mathrm{2}} \mathrm{20}}\right)\:\left(\frac{\mathrm{1}−\mathrm{tan}^{\mathrm{2}} \mathrm{20}}{\mathrm{tan20}}\right)=\mathrm{4tan10}\:\mathrm{proved}.\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$ \\ $$

Commented by Tony Lin last updated on 03/Jul/19

tan70−tan20−2tan40 =((sin70)/(cos70))−((sin20)/(cos20))−((2sin40)/(cos40)) =((sin50)/(cos70cos20))−((2sin40)/(cos40)) =((sin50cos40−2sin40cos70cos20)/(cos70cos20cos40)) =((sin50cos40−sin40(cos90+cos50))/((1/2)(cos90+cos50)cos40)) =((sin50cos40−cos50sin40)/((1/4)(cos90+cos10))) =((4sin10)/(cos10)) =4tan10

$${tan}\mathrm{70}−{tan}\mathrm{20}−\mathrm{2}{tan}\mathrm{40} \\ $$$$=\frac{{sin}\mathrm{70}}{{cos}\mathrm{70}}−\frac{{sin}\mathrm{20}}{{cos}\mathrm{20}}−\frac{\mathrm{2}{sin}\mathrm{40}}{{cos}\mathrm{40}} \\ $$$$=\frac{{sin}\mathrm{50}}{{cos}\mathrm{70}{cos}\mathrm{20}}−\frac{\mathrm{2}{sin}\mathrm{40}}{{cos}\mathrm{40}} \\ $$$$=\frac{{sin}\mathrm{50}{cos}\mathrm{40}−\mathrm{2}{sin}\mathrm{40}{cos}\mathrm{70}{cos}\mathrm{20}}{{cos}\mathrm{70}{cos}\mathrm{20}{cos}\mathrm{40}} \\ $$$$=\frac{{sin}\mathrm{50}{cos}\mathrm{40}−{sin}\mathrm{40}\left({cos}\mathrm{90}+{cos}\mathrm{50}\right)}{\frac{\mathrm{1}}{\mathrm{2}}\left({cos}\mathrm{90}+{cos}\mathrm{50}\right){cos}\mathrm{40}} \\ $$$$=\frac{{sin}\mathrm{50}{cos}\mathrm{40}−{cos}\mathrm{50}{sin}\mathrm{40}}{\frac{\mathrm{1}}{\mathrm{4}}\left({cos}\mathrm{90}+{cos}\mathrm{10}\right)} \\ $$$$=\frac{\mathrm{4}{sin}\mathrm{10}}{{cos}\mathrm{10}} \\ $$$$=\mathrm{4}{tan}\mathrm{10} \\ $$

Commented by Prithwish sen last updated on 03/Jul/19

Excellent sir

$$\mathrm{Excellent}\:\mathrm{sir} \\ $$

Answered by Hope last updated on 03/Jul/19

((sin70)/(cos70))−((sin20)/(cos20))−((2sin40)/(cos40)) =((sin(70−20))/(cos70cos20))−((2sin40)/(cos40)) =((2sin50)/(cos90+cos50))−((2sin40)/(cos40)) =((2sin50)/(cos50))−((2sin40)/(cos40)) =((2(sin50cos40−cos50sin40))/(2cos50cos40))×2 =((4sin10)/(cos90+cos10)) =4tan10

$$\frac{{sin}\mathrm{70}}{{cos}\mathrm{70}}−\frac{{sin}\mathrm{20}}{{cos}\mathrm{20}}−\frac{\mathrm{2}{sin}\mathrm{40}}{{cos}\mathrm{40}} \\ $$$$=\frac{{sin}\left(\mathrm{70}−\mathrm{20}\right)}{{cos}\mathrm{70}{cos}\mathrm{20}}−\frac{\mathrm{2}{sin}\mathrm{40}}{{cos}\mathrm{40}} \\ $$$$=\frac{\mathrm{2}{sin}\mathrm{50}}{{cos}\mathrm{90}+{cos}\mathrm{50}}−\frac{\mathrm{2}{sin}\mathrm{40}}{{cos}\mathrm{40}} \\ $$$$=\frac{\mathrm{2}{sin}\mathrm{50}}{{cos}\mathrm{50}}−\frac{\mathrm{2}{sin}\mathrm{40}}{{cos}\mathrm{40}} \\ $$$$=\frac{\mathrm{2}\left({sin}\mathrm{50}{cos}\mathrm{40}−{cos}\mathrm{50}{sin}\mathrm{40}\right)}{\mathrm{2}{cos}\mathrm{50}{cos}\mathrm{40}}×\mathrm{2} \\ $$$$=\frac{\mathrm{4}{sin}\mathrm{10}}{{cos}\mathrm{90}+{cos}\mathrm{10}} \\ $$$$=\mathrm{4}{tan}\mathrm{10} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

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