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