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Question Number 17420 by sushmitak last updated on 05/Jul/17

tan^6 (π/9)−33tan^4 (π/9)+27tan^2 (π/9)=?

$$\mathrm{tan}^{\mathrm{6}} \frac{\pi}{\mathrm{9}}−\mathrm{33tan}^{\mathrm{4}} \frac{\pi}{\mathrm{9}}+\mathrm{27tan}^{\mathrm{2}} \frac{\pi}{\mathrm{9}}=? \\ $$

Answered by Tinkutara last updated on 05/Jul/17

tan 3((π/9)) = (√3)  (√3) = ((3 tan (π/9) − tan^3  (π/9))/(1 − 3 tan^2  (π/9)))  3(1 + 9 tan^4  (π/9) − 6 tan^2  (π/9)) =  tan^2  (π/9) (9 + tan^4  (π/9) − 6 tan^2  (π/9))  On rearranging, we get  tan^6  (π/9) − 33 tan^4  (π/9) + 27 tan^2  (π/9) = 3

$$\mathrm{tan}\:\mathrm{3}\left(\frac{\pi}{\mathrm{9}}\right)\:=\:\sqrt{\mathrm{3}} \\ $$$$\sqrt{\mathrm{3}}\:=\:\frac{\mathrm{3}\:\mathrm{tan}\:\frac{\pi}{\mathrm{9}}\:−\:\mathrm{tan}^{\mathrm{3}} \:\frac{\pi}{\mathrm{9}}}{\mathrm{1}\:−\:\mathrm{3}\:\mathrm{tan}^{\mathrm{2}} \:\frac{\pi}{\mathrm{9}}} \\ $$$$\mathrm{3}\left(\mathrm{1}\:+\:\mathrm{9}\:\mathrm{tan}^{\mathrm{4}} \:\frac{\pi}{\mathrm{9}}\:−\:\mathrm{6}\:\mathrm{tan}^{\mathrm{2}} \:\frac{\pi}{\mathrm{9}}\right)\:= \\ $$$$\mathrm{tan}^{\mathrm{2}} \:\frac{\pi}{\mathrm{9}}\:\left(\mathrm{9}\:+\:\mathrm{tan}^{\mathrm{4}} \:\frac{\pi}{\mathrm{9}}\:−\:\mathrm{6}\:\mathrm{tan}^{\mathrm{2}} \:\frac{\pi}{\mathrm{9}}\right) \\ $$$$\mathrm{On}\:\mathrm{rearranging},\:\mathrm{we}\:\mathrm{get} \\ $$$$\mathrm{tan}^{\mathrm{6}} \:\frac{\pi}{\mathrm{9}}\:−\:\mathrm{33}\:\mathrm{tan}^{\mathrm{4}} \:\frac{\pi}{\mathrm{9}}\:+\:\mathrm{27}\:\mathrm{tan}^{\mathrm{2}} \:\frac{\pi}{\mathrm{9}}\:=\:\mathrm{3} \\ $$

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