Question Number 54537 by hsbebeb last updated on 05/Feb/19
$${tan}^{\mathrm{2}} \mathrm{20}+{tan}^{\mathrm{2}} \mathrm{40}+{tan}^{\mathrm{2}} \mathrm{80}=\mathrm{33}\:\:\: \\ $$$$\left({please}\:{solve}\:{this}\:{and}\:{i}\:{want}\:{to}\:{know}\:\right. \\ $$$$\left.{that}\:{which}\:{standard}\:{that}\:{question}\:{belongs}?\right) \\ $$$${help}\:{needed} \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 05/Feb/19
Commented by tanmay.chaudhury50@gmail.com last updated on 05/Feb/19
$${download}\:{this}\:{book}\:{free}\:{from}\:{website}\: \\ $$$${archive}.{org} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 05/Feb/19
![θ=20^o 9θ=π tan9θ=tanπ=0 formula tan(nθ)=((s_1 −s_3 +s_5 −s_7 +...)/(1−s_2 +s_4 −...)) tan9θ=0=((9c_1 tanθ−9c_3 tan^3 θ+9c_5 tan^5 θ−9c_7 tan^7 θ+9c_9 tan^9 θ)/D_r ) tanθ=x 9x−((9×8×7)/(3×2))x^3 +((9×8×7×6)/(4×3×2))x^5 −((9×8)/2)x^7 +x^9 =0 9−84x^2 +14×9x^4 −36x^6 +x^8 =0 roots of this eqn are tanθ,tan2θ,...[θ=20^o ...] y=x^2 y^4 −36y^2 +126y^2 −84y+9=0 roots of this eqn tan^2 θ , tan^2 2θ,tan^2 3θ,tan^2 4θ sum of roots=−(−36)=36 tan^2 20+tan^2 40+tan^2 60+tan^2 80=36 so tan^2 20+tan^2 40+tan^2 80=36−((√3) )^2 =33](https://www.tinkutara.com/question/Q54548.png)
$$\theta=\mathrm{20}^{{o}} \\ $$$$\mathrm{9}\theta=\pi \\ $$$${tan}\mathrm{9}\theta={tan}\pi=\mathrm{0} \\ $$$${formula}\:\:\:{tan}\left({n}\theta\right)=\frac{{s}_{\mathrm{1}} −{s}_{\mathrm{3}} +{s}_{\mathrm{5}} −{s}_{\mathrm{7}} +…}{\mathrm{1}−{s}_{\mathrm{2}} +{s}_{\mathrm{4}} −…} \\ $$$${tan}\mathrm{9}\theta=\mathrm{0}=\frac{\mathrm{9}{c}_{\mathrm{1}} {tan}\theta−\mathrm{9}{c}_{\mathrm{3}} {tan}^{\mathrm{3}} \theta+\mathrm{9}{c}_{\mathrm{5}} {tan}^{\mathrm{5}} \theta−\mathrm{9}{c}_{\mathrm{7}} {tan}^{\mathrm{7}} \theta+\mathrm{9}{c}_{\mathrm{9}} {tan}^{\mathrm{9}} \theta}{{D}_{{r}} } \\ $$$${tan}\theta={x} \\ $$$$\mathrm{9}{x}−\frac{\mathrm{9}×\mathrm{8}×\mathrm{7}}{\mathrm{3}×\mathrm{2}}{x}^{\mathrm{3}} +\frac{\mathrm{9}×\mathrm{8}×\mathrm{7}×\mathrm{6}}{\mathrm{4}×\mathrm{3}×\mathrm{2}}{x}^{\mathrm{5}} −\frac{\mathrm{9}×\mathrm{8}}{\mathrm{2}}{x}^{\mathrm{7}} +{x}^{\mathrm{9}} =\mathrm{0} \\ $$$$\mathrm{9}−\mathrm{84}{x}^{\mathrm{2}} +\mathrm{14}×\mathrm{9}{x}^{\mathrm{4}} −\mathrm{36}{x}^{\mathrm{6}} +{x}^{\mathrm{8}} =\mathrm{0} \\ $$$${roots}\:{of}\:{this}\:{eqn}\:{are}\:{tan}\theta,{tan}\mathrm{2}\theta,…\left[\theta=\mathrm{20}^{{o}} …\right] \\ $$$${y}={x}^{\mathrm{2}} \\ $$$${y}^{\mathrm{4}} −\mathrm{36}{y}^{\mathrm{2}} +\mathrm{126}{y}^{\mathrm{2}} −\mathrm{84}{y}+\mathrm{9}=\mathrm{0} \\ $$$${roots}\:{of}\:{this}\:{eqn}\:{tan}^{\mathrm{2}} \theta\:,\:{tan}^{\mathrm{2}} \mathrm{2}\theta,{tan}^{\mathrm{2}} \mathrm{3}\theta,{tan}^{\mathrm{2}} \mathrm{4}\theta \\ $$$${sum}\:{of}\:{roots}=−\left(−\mathrm{36}\right)=\mathrm{36} \\ $$$${tan}^{\mathrm{2}} \mathrm{20}+{tan}^{\mathrm{2}} \mathrm{40}+{tan}^{\mathrm{2}} \mathrm{60}+{tan}^{\mathrm{2}} \mathrm{80}=\mathrm{36} \\ $$$${so}\:{tan}^{\mathrm{2}} \mathrm{20}+{tan}^{\mathrm{2}} \mathrm{40}+{tan}^{\mathrm{2}} \mathrm{80}=\mathrm{36}−\left(\sqrt{\mathrm{3}}\:\right)^{\mathrm{2}} =\mathrm{33} \\ $$$$ \\ $$