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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

$$\mathrm{If}\:\:\:\mathrm{for}\:\:\mathrm{0}<{x}<\frac{\pi}{\mathrm{2}}\:, \\ $$ $$\:{y}={exp}\left[\left(\mathrm{sin}^{\mathrm{2}} {x}+\mathrm{sin}^{\mathrm{4}} {x}+\mathrm{sin}^{\mathrm{6}} {x}+...\infty\right)\mathrm{log}_{{e}} \mathrm{2}\right] \\ $$ $$\mathrm{is}\:\mathrm{a}\:\mathrm{zero}\:\mathrm{of}\:\mathrm{the}\:\mathrm{quadratic}\:\mathrm{equation}\: \\ $$ $${x}^{\mathrm{2}} −\mathrm{9}{x}+\mathrm{8}=\mathrm{0},\:\mathrm{then}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\frac{\mathrm{sin}\:{x}+\mathrm{cos}\:{x}}{\mathrm{sin}\:{x}−\mathrm{cos}\:{x}}\:\mathrm{is} \\ $$

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

$${y}={e}^{\left[\left({sin}^{\mathrm{2}} {x}+{sin}^{\mathrm{4}} {x}+...\infty\right){ln}\mathrm{2}\right]} \\ $$ $${now}\:{s}={sin}^{\mathrm{2}} {x}+{sin}^{\mathrm{4}} {x}+...\infty \\ $$ $${s}=\frac{{sin}^{\mathrm{2}} {x}}{\mathrm{1}−{sin}^{\mathrm{2}} {x}}={tan}^{\mathrm{2}} {x} \\ $$ $${y}={e}^{{tan}^{\mathrm{2}} {x}×{ln}\mathrm{2}} \\ $$ $${x}^{\mathrm{2}} −\mathrm{9}{x}+\mathrm{8}=\mathrm{0} \\ $$ $$\left({x}−\mathrm{1}\right)\left({x}−\mathrm{8}\right)=\mathrm{0} \\ $$ $${so}\:{value}\:{of}\:{y}\:{is}\:{either}=\mathrm{1}\:\:\:{or}\:\:\mathrm{8} \\ $$ $$\mathrm{1}={e}^{{tan}^{\mathrm{2}} {x}×{ln}\mathrm{2}} \\ $$ $${e}^{\mathrm{0}} ={e}^{{tan}^{\mathrm{2}} {x}×{ln}\mathrm{2}} \\ $$ $$\boldsymbol{{but}}\:\boldsymbol{{tanx}}\neq\mathrm{0} \\ $$ $${so}\:\mathrm{8}={e}^{{tan}^{\mathrm{2}} {x}×{ln}\mathrm{2}} \\ $$ $${tan}^{\mathrm{2}} {x}×{ln}\mathrm{2}={ln}\mathrm{8} \\ $$ $${tan}^{\mathrm{2}} {x}×{ln}\mathrm{2}=\mathrm{3}{ln}\mathrm{2} \\ $$ $${tanx}=\sqrt{\mathrm{3}}\: \\ $$ $$\frac{{sinx}+{cosx}}{{sinx}−{cosx}} \\ $$ $$=\frac{{tanx}+\mathrm{1}}{{tanx}−\mathrm{1}}\rightarrow\frac{\sqrt{\mathrm{3}}\:+\mathrm{1}}{\sqrt{\mathrm{3}}\:−\mathrm{1}} \\ $$ $$\frac{\mathrm{3}+\mathrm{2}\sqrt{\mathrm{3}}\:+\mathrm{1}}{\mathrm{3}−\mathrm{1}} \\ $$ $$=\frac{\mathrm{4}+\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$ $$=\mathrm{2}+\sqrt{\mathrm{3}}\:\boldsymbol{{answer}} \\ $$ $$ \\ $$

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