Question Number 56415 by gunawan last updated on 16/Mar/19
![If for 0<x<(π/2) , y=exp[(sin^2 x+sin^4 x+sin^6 x+...∞)log_e 2] is a zero of the quadratic equation x^2 −9x+8=0, then the value of((sin x+cos x)/(sin x−cos x)) is](https://www.tinkutara.com/question/Q56415.png)
$$\mathrm{If}\:\:\:\mathrm{for}\:\:\mathrm{0}<{x}<\frac{\pi}{\mathrm{2}}\:, \\ $$$$\:{y}={exp}\left[\left(\mathrm{sin}^{\mathrm{2}} {x}+\mathrm{sin}^{\mathrm{4}} {x}+\mathrm{sin}^{\mathrm{6}} {x}+…\infty\right)\mathrm{log}_{{e}} \mathrm{2}\right] \\ $$$$\mathrm{is}\:\mathrm{a}\:\mathrm{zero}\:\mathrm{of}\:\mathrm{the}\:\mathrm{quadratic}\:\mathrm{equation}\: \\ $$$${x}^{\mathrm{2}} −\mathrm{9}{x}+\mathrm{8}=\mathrm{0},\:\mathrm{then}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\frac{\mathrm{sin}\:{x}+\mathrm{cos}\:{x}}{\mathrm{sin}\:{x}−\mathrm{cos}\:{x}}\:\mathrm{is} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 16/Mar/19
![y=e^([(sin^2 x+sin^4 x+...∞)ln2]) now s=sin^2 x+sin^4 x+...∞ s=((sin^2 x)/(1−sin^2 x))=tan^2 x y=e^(tan^2 x×ln2) x^2 −9x+8=0 (x−1)(x−8)=0 so value of y is either=1 or 8 1=e^(tan^2 x×ln2) e^0 =e^(tan^2 x×ln2) but tanx≠0 so 8=e^(tan^2 x×ln2) tan^2 x×ln2=ln8 tan^2 x×ln2=3ln2 tanx=(√3) ((sinx+cosx)/(sinx−cosx)) =((tanx+1)/(tanx−1))→(((√3) +1)/( (√3) −1)) ((3+2(√3) +1)/(3−1)) =((4+2(√3))/2) =2+(√3) answer](https://www.tinkutara.com/question/Q56446.png)
$${y}={e}^{\left[\left({sin}^{\mathrm{2}} {x}+{sin}^{\mathrm{4}} {x}+…\infty\right){ln}\mathrm{2}\right]} \\ $$$${now}\:{s}={sin}^{\mathrm{2}} {x}+{sin}^{\mathrm{4}} {x}+…\infty \\ $$$${s}=\frac{{sin}^{\mathrm{2}} {x}}{\mathrm{1}−{sin}^{\mathrm{2}} {x}}={tan}^{\mathrm{2}} {x} \\ $$$${y}={e}^{{tan}^{\mathrm{2}} {x}×{ln}\mathrm{2}} \\ $$$${x}^{\mathrm{2}} −\mathrm{9}{x}+\mathrm{8}=\mathrm{0} \\ $$$$\left({x}−\mathrm{1}\right)\left({x}−\mathrm{8}\right)=\mathrm{0} \\ $$$${so}\:{value}\:{of}\:{y}\:{is}\:{either}=\mathrm{1}\:\:\:{or}\:\:\mathrm{8} \\ $$$$\mathrm{1}={e}^{{tan}^{\mathrm{2}} {x}×{ln}\mathrm{2}} \\ $$$${e}^{\mathrm{0}} ={e}^{{tan}^{\mathrm{2}} {x}×{ln}\mathrm{2}} \\ $$$$\boldsymbol{{but}}\:\boldsymbol{{tanx}}\neq\mathrm{0} \\ $$$${so}\:\mathrm{8}={e}^{{tan}^{\mathrm{2}} {x}×{ln}\mathrm{2}} \\ $$$${tan}^{\mathrm{2}} {x}×{ln}\mathrm{2}={ln}\mathrm{8} \\ $$$${tan}^{\mathrm{2}} {x}×{ln}\mathrm{2}=\mathrm{3}{ln}\mathrm{2} \\ $$$${tanx}=\sqrt{\mathrm{3}}\: \\ $$$$\frac{{sinx}+{cosx}}{{sinx}−{cosx}} \\ $$$$=\frac{{tanx}+\mathrm{1}}{{tanx}−\mathrm{1}}\rightarrow\frac{\sqrt{\mathrm{3}}\:+\mathrm{1}}{\:\sqrt{\mathrm{3}}\:−\mathrm{1}} \\ $$$$\frac{\mathrm{3}+\mathrm{2}\sqrt{\mathrm{3}}\:+\mathrm{1}}{\mathrm{3}−\mathrm{1}} \\ $$$$=\frac{\mathrm{4}+\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$$=\mathrm{2}+\sqrt{\mathrm{3}}\:\boldsymbol{{answer}} \\ $$$$ \\ $$