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Question Number 18731 by juniorjoseph679@gmail.com last updated on 29/Jul/17

tan ((2π)/5) − tan (π/(15)) − (√3) tan ((2π)/5) tan (π/(15)) =

$$\mathrm{tan}\:\frac{\mathrm{2}\pi}{\mathrm{5}}\:−\:\:\mathrm{tan}\:\frac{\pi}{\mathrm{15}}\:−\:\sqrt{\mathrm{3}}\:\mathrm{tan}\:\frac{\mathrm{2}\pi}{\mathrm{5}}\:\mathrm{tan}\:\frac{\pi}{\mathrm{15}}\:= \\ $$

Answered by 32 last updated on 29/Jul/17

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Answered by Tinkutara last updated on 29/Jul/17

tan (π/3) = (√3) = tan (((2π)/5) − (π/(15))) (√3) = ((tan ((2π)/5) − tan (π/(15)))/(1 + tan ((2π)/5) tan (π/(15)))) tan ((2π)/5) − tan (π/(15)) = (√3) + (√3) tan ((2π)/5) tan (π/(15)) tan ((2π)/5) − tan (π/(15)) − (√3) tan ((2π)/5) tan (π/(15)) = (√3)

$$\mathrm{tan}\:\frac{\pi}{\mathrm{3}}\:=\:\sqrt{\mathrm{3}}\:=\:\mathrm{tan}\:\left(\frac{\mathrm{2}\pi}{\mathrm{5}}\:−\:\frac{\pi}{\mathrm{15}}\right) \\ $$$$\sqrt{\mathrm{3}}\:=\:\frac{\mathrm{tan}\:\frac{\mathrm{2}\pi}{\mathrm{5}}\:−\:\mathrm{tan}\:\frac{\pi}{\mathrm{15}}}{\mathrm{1}\:+\:\mathrm{tan}\:\frac{\mathrm{2}\pi}{\mathrm{5}}\:\mathrm{tan}\:\frac{\pi}{\mathrm{15}}} \\ $$$$\mathrm{tan}\:\frac{\mathrm{2}\pi}{\mathrm{5}}\:−\:\mathrm{tan}\:\frac{\pi}{\mathrm{15}}\:=\:\sqrt{\mathrm{3}}\:+\:\sqrt{\mathrm{3}}\:\mathrm{tan}\:\frac{\mathrm{2}\pi}{\mathrm{5}}\:\mathrm{tan}\:\frac{\pi}{\mathrm{15}} \\ $$$$\mathrm{tan}\:\frac{\mathrm{2}\pi}{\mathrm{5}}\:−\:\mathrm{tan}\:\frac{\pi}{\mathrm{15}}\:−\:\sqrt{\mathrm{3}}\:\mathrm{tan}\:\frac{\mathrm{2}\pi}{\mathrm{5}}\:\mathrm{tan}\:\frac{\pi}{\mathrm{15}}\:=\:\sqrt{\mathrm{3}} \\ $$

Answered by 32 last updated on 29/Jul/17

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