Question-48367

Question Number 48367 by behi83417@gmail.com last updated on 22/Nov/18

Answered by tanmay.chaudhury50@gmail.com last updated on 23/Nov/18

xy−1=(√3) (x+y) (1/( (√3)))=((x+y)/(xy−1)) ((x+y)/(1−xy))=−(1/( (√3))) tana=x tanb=y tanc=z tan(a+b)=−(1/( (√3))) tan(b+c)=−(1/( (√3))) tan(a+c)=−(1/( (√3))) tan(a+b)=tan(b+c)=tan(a+c) so a=b=c so tana=tanb=tanc x=y=z ((x+x)/(1−x×x))=−(1/( (√3))) 2(√3) x =−1(1−x^2 ) 2(√3) x+1−x^2 =0 x^2 −2(√3) x−1=0 x=((2(√3) ±(√(12+4)) )/2)=(√3) ±2 x=y=z=(√3) ±2

$${xy}−\mathrm{1}=\sqrt{\mathrm{3}}\:\left({x}+{y}\right) \\ $$$$\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}=\frac{{x}+{y}}{{xy}−\mathrm{1}} \\ $$$$\frac{{x}+{y}}{\mathrm{1}−{xy}}=−\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}} \\ $$$${tana}={x}\:\:\:{tanb}={y}\:\:\:\:{tanc}={z} \\ $$$${tan}\left({a}+{b}\right)=−\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}} \\ $$$${tan}\left({b}+{c}\right)=−\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}} \\ $$$${tan}\left({a}+{c}\right)=−\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}} \\ $$$${tan}\left({a}+{b}\right)={tan}\left({b}+{c}\right)={tan}\left({a}+{c}\right)\:\:{so} \\ $$$${a}={b}={c}\:\:{so} \\ $$$$ \\ $$$${tana}={tanb}={tanc} \\ $$$${x}={y}={z} \\ $$$$\frac{{x}+{x}}{\mathrm{1}−{x}×{x}}=−\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}} \\ $$$$\mathrm{2}\sqrt{\mathrm{3}}\:{x}\:=−\mathrm{1}\left(\mathrm{1}−{x}^{\mathrm{2}} \right) \\ $$$$\mathrm{2}\sqrt{\mathrm{3}}\:{x}+\mathrm{1}−{x}^{\mathrm{2}} \:=\mathrm{0} \\ $$$${x}^{\mathrm{2}} −\mathrm{2}\sqrt{\mathrm{3}}\:{x}−\mathrm{1}=\mathrm{0} \\ $$$${x}=\frac{\mathrm{2}\sqrt{\mathrm{3}}\:\pm\sqrt{\mathrm{12}+\mathrm{4}}\:}{\mathrm{2}}=\sqrt{\mathrm{3}}\:\pm\mathrm{2} \\ $$$${x}={y}={z}=\sqrt{\mathrm{3}}\:\pm\mathrm{2} \\ $$$$ \\ $$

Commented by behi83417@gmail.com last updated on 23/Nov/18

$${very}\:{beautiful}\:{method}\:!{god}\:{bless}\:{you}\:{sir}. \\ $$

Commented by tanmay.chaudhury50@gmail.com last updated on 23/Nov/18

$${thank}\:{you}\:{sir}… \\ $$

Answered by MJS last updated on 22/Nov/18

(1) x=((1+y(√3))/(y−(√3))) (2) z=((1+y(√3))/(y−(√3)))=x (3) (((1+y(√3))^2 )/((y−(√3))^2 ))=2(√3)((1+y(√3))/(y−(√3)))+1 ⇒ y^2 −2(√3)y−1=0 ⇒ y=±2+(√3) x=y=x=±2+(√3)

$$\left(\mathrm{1}\right)\:\:{x}=\frac{\mathrm{1}+{y}\sqrt{\mathrm{3}}}{{y}−\sqrt{\mathrm{3}}} \\ $$$$\left(\mathrm{2}\right)\:\:{z}=\frac{\mathrm{1}+{y}\sqrt{\mathrm{3}}}{{y}−\sqrt{\mathrm{3}}}={x} \\ $$$$\left(\mathrm{3}\right)\:\:\frac{\left(\mathrm{1}+{y}\sqrt{\mathrm{3}}\right)^{\mathrm{2}} }{\left({y}−\sqrt{\mathrm{3}}\right)^{\mathrm{2}} }=\mathrm{2}\sqrt{\mathrm{3}}\frac{\mathrm{1}+{y}\sqrt{\mathrm{3}}}{{y}−\sqrt{\mathrm{3}}}+\mathrm{1} \\ $$$$\Rightarrow\:{y}^{\mathrm{2}} −\mathrm{2}\sqrt{\mathrm{3}}{y}−\mathrm{1}=\mathrm{0}\:\Rightarrow\:{y}=\pm\mathrm{2}+\sqrt{\mathrm{3}} \\ $$$${x}={y}={x}=\pm\mathrm{2}+\sqrt{\mathrm{3}} \\ $$

Commented by behi83417@gmail.com last updated on 23/Nov/18

$${so}\:{fast}!{thanks}\:{in}\:{advance}\:{sir}. \\ $$

Commented by MJS last updated on 23/Nov/18

if we see that x=y=z at start we can directly go to x^2 =2(√3)x+1

$$\mathrm{if}\:\mathrm{we}\:\mathrm{see}\:\mathrm{that}\:{x}={y}={z}\:\mathrm{at}\:\mathrm{start}\:\mathrm{we}\:\mathrm{can}\:\mathrm{directly} \\ $$$$\mathrm{go}\:\mathrm{to}\:{x}^{\mathrm{2}} =\mathrm{2}\sqrt{\mathrm{3}}{x}+\mathrm{1} \\ $$

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