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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

$$xy-1=\sqrt{3}(x+y)$$$$\frac{1}{\sqrt{3}}=\frac{x+y}{xy-1}$$$$\frac{x+y}{1-xy}=-\frac{1}{\sqrt{3}}$$$$tana=xtanb=ytanc=z$$$$tan(a+b)=-\frac{1}{\sqrt{3}}$$$$tan(b+c)=-\frac{1}{\sqrt{3}}$$$$tan(a+c)=-\frac{1}{\sqrt{3}}$$$$tan(a+b)=tan(b+c)=tan(a+c)so$$$$a=b=cso$$$$$$$$tana=tanb=tanc$$$$x=y=z$$$$\frac{x+x}{1-x\times x}=-\frac{1}{\sqrt{3}}$$$$2\sqrt{3}x=-1(1-x^2)$$$$2\sqrt{3}x+1-x^2=0$$$$x^2-2\sqrt{3}x-1=0$$$$x=\frac{2\sqrt{3}\pm \sqrt{12+4}}{2}=\sqrt{3}\pm 2$$$$x=y=z=\sqrt{3}\pm 2$$$$$$

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

$$verybeautifulmethod!godblessyousir.$$

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

$$thankyousir...$$

Answered by MJS last updated on 22/Nov/18

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

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

$$sofast!thanksinadvancesir.$$

Commented by MJS last updated on 23/Nov/18

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

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