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I wanted to know how can i prove that if

$xy+yz+zx=1$, then

$$ \frac{x}{1+x^2}+\frac{y}{1+y^2}+\frac{z}{1+z^2} = \frac{2}{\sqrt{(1+x^2)(1+y^2)(1+z^2)}}$$

I did let $x=\tan A$, $y=\tan B$, $z=\tan C$

given $xy+yz+zx =1$ we have $\tan A \tan B+ \tan B \tan C+\tan C \tan A=1$

$\tan C(\tan A+\tan B)=1-\tan A \tan B$, or $\tan(A+B)=\tan(\pi/2 -C)$ we have $A+B+C=\pi/2$.

what to do now?

Any help appreciated.


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So your question is whether this problem is tough? Because your title indicates exactly that (it also indicates, by duplication of the question mark, that you don't really believe it). – celtschk Jul 21 '13 at 9:55
@celtschk: Maybe the question is whether it is algebric. – Chris Eagle Jul 21 '13 at 11:21
up vote 7 down vote accepted


$$\frac x{1+x^2}=\frac {\tan A}{1+\tan^2A}=\frac{2\sin A\cos A}2=\frac{\sin2A}2$$

Now , $$\sin2A+\sin2B+\sin2C=2\sin(A+B)\cos(A-B)+2\sin C\cos C$$

$$=2\sin\left(\frac\pi2-C\right)\cos(A-B)+2\sin C\cos C$$

$$=2\cos C\{\cos(A-B)+\cos(A+B)\}$$ as $\sin C=\sin\{\frac\pi2-(A+B)\}=\cos(A+B)$

$$\implies \sin2A+\sin2B+\sin2C=2\cos C\cdot2\cos A\cos B$$

and $$\frac1{\sqrt{1+x^2}}=\frac1{\sec A}=\cos A$$

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do you people even blink, before answering o_o. – SHOBHIT GAUTAM Jul 21 '13 at 4:59
@Shobhit, not sure if I clearly understood your statement. But have you noticed $\tan(A+B)=\tan\left(\frac\pi2-C\right)\implies A+B=n\pi+\frac\pi2-C$ where $n$ is any integer. The above proof hold if $n=2m$(even). What about $n=2m+1$(odd)? – lab bhattacharjee Jul 21 '13 at 5:09
@Shobhit: This is a really cool problem. May I know the name of the book you are studying from? – Prism Jul 21 '13 at 6:11
@Shobhit, if $n=2m+1,$ $$\tan\left(A+B\right)=\tan\{(2m+1)\pi+\frac\pi2-C\}=\tan\left(\frac\pi2-C\right‌​)=\cot C,$$ but $$\sin\left(A+B\right)=\sin\{(2m+1)\pi+\frac\pi2-C\}=-\sin\left(\frac\pi2-C\righ‌​t)=-\cos C$$ and $$\cos\left(A+B\right)=\cos\{(2m+1)\pi+\frac\pi2-C\}=-\cos\left(\frac\pi2-C\righ‌​t)=-\sin C$$ – lab bhattacharjee Jul 21 '13 at 6:33
@labbhattacharjee the maths written is not clear mind to take a look. – SHOBHIT GAUTAM Jul 21 '13 at 7:33

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