Question

If $xy+yz+zx=1$, the prove that $\frac{x}{1+x^2}+\frac{y}{1+y^2}+\frac{z}{1+z^2}=\frac{2}{\left[\right(1=x^2\left)\right(1+y^2\left)\right(1+z^2){]}^{1/2}}$.

Answer 1

Put $x=\tan \alpha$ etc.
so that $1-S_2=0$, given.
$\therefore \tan (\alpha +\beta +\gamma )=\frac{S_1-S_3}{1-S_2}=\infty$
$\therefore \alpha +\beta +\gamma =\frac{\pi }{2}$
Also $\frac{x}{1+x^2}=\frac{\tan \alpha }{se{c}^2\alpha }=\sin \alpha cos\alpha =\frac{1}{2}\sin 2\alpha$
Hence we have to prove that
$\frac{1}{2}[\sin 2\alpha +\sin 2\beta +\sin 2\gamma ]=\frac{2}{sec\alpha sec\beta sec\gamma }$
or $\sin 2\alpha +\sin 2\beta +\sin 2\gamma =4\cos \alpha \cos \beta \cos \gamma$
when $\alpha +\beta +\gamma =\pi /2$
or $\sin (\beta -\gamma )=\sin \left\{\right(\pi /2)-\alpha \}=\cos \alpha$
Now above relation can be easily proved.