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Question-210677




Question Number 210677 by Safojon last updated on 16/Aug/24
Answered by BHOOPENDRA last updated on 16/Aug/24
θ=((kπ)/(7 )),7θ=kπ (where k=1,2,3,4....n)  7θ=π (k=1)  4θ+3θ=π  tg4θ=−tg(3θ) let tg=t  ⇒ ((4t−4t^3 )/(1−6t^2 +t^4 ))=((−3t−t^3 )/(1−3t^2 ))  =t^6 −21t^4 +35t^2 −7=0   eq(1)  = This is a cubic equation in t^2  that  is in tg^2 θ.   the root of the eq is tg^2 (π/7),tg^2 ((2π)/7)  tg^2 ((3π)/7)  So sum of the root =−(((−21))/1)=21  tg^2 (π/7)+tg^2 ((2π)/7)+tg^2 ((3π)/7)=21
$$\theta=\frac{{k}\pi}{\mathrm{7}\:},\mathrm{7}\theta={k}\pi\:\left({where}\:{k}=\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{4}….{n}\right) \\ $$$$\mathrm{7}\theta=\pi\:\left({k}=\mathrm{1}\right) \\ $$$$\mathrm{4}\theta+\mathrm{3}\theta=\pi \\ $$$${tg}\mathrm{4}\theta=−{tg}\left(\mathrm{3}\theta\right)\:{let}\:{tg}={t} \\ $$$$\Rightarrow\:\frac{\mathrm{4}{t}−\mathrm{4}{t}^{\mathrm{3}} }{\mathrm{1}−\mathrm{6}{t}^{\mathrm{2}} +{t}^{\mathrm{4}} }=\frac{−\mathrm{3}{t}−{t}^{\mathrm{3}} }{\mathrm{1}−\mathrm{3}{t}^{\mathrm{2}} } \\ $$$$={t}^{\mathrm{6}} −\mathrm{21}{t}^{\mathrm{4}} +\mathrm{35}{t}^{\mathrm{2}} −\mathrm{7}=\mathrm{0}\:\:\:{eq}\left(\mathrm{1}\right) \\ $$$$=\:{This}\:{is}\:{a}\:{cubic}\:{equation}\:{in}\:{t}^{\mathrm{2}} \:{that} \\ $$$${is}\:{in}\:{tg}^{\mathrm{2}} \theta. \\ $$$$\:{the}\:{root}\:{of}\:{the}\:{eq}\:{is}\:{tg}^{\mathrm{2}} \frac{\pi}{\mathrm{7}},{tg}^{\mathrm{2}} \frac{\mathrm{2}\pi}{\mathrm{7}} \\ $$$${tg}^{\mathrm{2}} \frac{\mathrm{3}\pi}{\mathrm{7}} \\ $$$${So}\:{sum}\:{of}\:{the}\:{root}\:=−\frac{\left(−\mathrm{21}\right)}{\mathrm{1}}=\mathrm{21} \\ $$$${tg}^{\mathrm{2}} \frac{\pi}{\mathrm{7}}+{tg}^{\mathrm{2}} \frac{\mathrm{2}\pi}{\mathrm{7}}+{tg}^{\mathrm{2}} \frac{\mathrm{3}\pi}{\mathrm{7}}=\mathrm{21} \\ $$$$ \\ $$$$ \\ $$

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