Question Number 39848 by ajfour last updated on 12/Jul/18
Answered by tanmay.chaudhury50@gmail.com last updated on 12/Jul/18
$${tan}\theta=\frac{{a}}{\mathrm{1}} \\ $$$${sin}\theta=\frac{\mathrm{1}}{\mathrm{1}+{a}} \\ $$$${tan}\theta=\frac{{a}}{\mathrm{1}}\:\:{so}\:{sin}\theta=\frac{{a}}{\sqrt{\mathrm{1}+{a}^{\mathrm{2}} }} \\ $$$$\frac{{a}}{\sqrt{\mathrm{1}+{a}^{\mathrm{2}} }}=\frac{\mathrm{1}}{{a}+\mathrm{1}} \\ $$$${a}^{\mathrm{2}} +{a}=\sqrt{\mathrm{1}+{a}^{\mathrm{2}} } \\ $$$${a}^{\mathrm{4}} +\mathrm{2}{a}^{\mathrm{3}} +{a}^{\mathrm{2}} ={a}^{\mathrm{2}} +\mathrm{1} \\ $$$${a}^{\mathrm{4}} +\mathrm{2}{a}^{\mathrm{3}} −\mathrm{1}=\mathrm{0} \\ $$$${solving}... \\ $$$${f}\left({a}\right)={a}^{\mathrm{4}} +\mathrm{2}{a}^{\mathrm{3}} −\mathrm{1} \\ $$$${f}\left(\mathrm{1}\right)=\mathrm{2}>\mathrm{0} \\ $$$${f}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)=\frac{\mathrm{1}}{\mathrm{16}}+\frac{\mathrm{2}}{\mathrm{8}}−\mathrm{1}=\frac{−\mathrm{11}}{\mathrm{16}} \\ $$$${since}\:{f}\left(\mathrm{1}\right)>\mathrm{0}\:{but}\:{f}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)<\mathrm{0}\:\:{that}\:{means}\: \\ $$$${in}\:{between}\:\frac{\mathrm{1}}{\mathrm{2}}\:{and}\:\mathrm{1}\:{the}\:{curve}\: \\ $$$${f}\left({a}\right)={a}^{\mathrm{4}} +\mathrm{2}{a}^{\mathrm{3}} −\mathrm{1}\:{change}\:{its}\:{sign}\:... \\ $$$${so}\:{value}\:{of}\:{a}\:{lies}\:{between}\:\left(\frac{\mathrm{1}}{\mathrm{2}}{and}\mathrm{1}\right) \\ $$$$\:\:\:\mathrm{1}>{a}>\frac{\mathrm{1}}{\mathrm{2}} \\ $$
Commented by ajfour last updated on 12/Jul/18
$${Thanks}\:{Tanmay}\:{Sir}. \\ $$
Answered by MrW3 last updated on 12/Jul/18
$$\frac{{a}}{\mathrm{1}}=\frac{\mathrm{1}}{\sqrt{\left(\mathrm{1}+{a}\right)^{\mathrm{2}} −\mathrm{1}^{\mathrm{2}} }} \\ $$$${a}\sqrt{{a}\left({a}+\mathrm{2}\right)}=\mathrm{1} \\ $$$${a}^{\mathrm{4}} +\mathrm{2}{a}^{\mathrm{3}} −\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow{a}\approx\mathrm{0}.\mathrm{7167} \\ $$
Commented by ajfour last updated on 12/Jul/18
$${Thanks}\:{MrW}\:{Sir}.\: \\ $$