Proving Given identity for sum of \( \arctan(a)+\arctan(b)+\arctan(c) \)

Exercise:

Show that, $\arctan \sqrt{\frac{x\left( x+y+z \right)}{yz}}+\arctan \sqrt{\frac{y\left( x+y+z \right)}{zx}}+\arctan \sqrt{\frac{z\left( x+y+z \right)}{xy}}=\pi $

Solution: Let $w=x+y+z>0$

So $L.H.S=\arctan \sqrt{\frac{xw}{yz}}+\arctan \sqrt{\frac{yw}{zx}}+\arctan \sqrt{\frac{zw}{xy}}$

Let $\theta =\arctan \sqrt{\frac{xw}{yz}}\,\,,\gamma =\arctan \sqrt{\frac{yw}{zx}}\,\,\And \,\,\,\varphi =\arctan \sqrt{\frac{zw}{xy}}$

Note that, $\tan \left( X+Y \right)=\frac{\tan X+\tan Y}{1-\tan X\tan Y}$

So $\tan \left( \theta +\gamma +\varphi  \right)=\frac{\tan \left( \theta +\gamma  \right)+\tan \varphi }{1-\tan \left( \theta +\gamma  \right)\tan \varphi }=\frac{\frac{\tan \theta +\tan \gamma }{1-\tan \theta \tan \gamma }+\tan \varphi }{1-\frac{\tan \theta +\tan \gamma }{1-\tan \theta \tan \gamma }\tan \varphi }$

$=\frac{\frac{\tan \theta +\tan \gamma +\tan \varphi \left( 1-\tan \theta \tan \gamma  \right)}{1-\tan \theta \tan \gamma }}{\frac{1-\tan \theta \tan \gamma -\left( \tan \theta +\tan \gamma  \right)\tan \varphi }{1-\tan \theta \tan \gamma }}=\frac{\tan \theta +\tan \gamma +\tan \varphi -\tan \varphi \tan \theta \tan \gamma }{1-\tan \theta \tan \gamma -\tan \theta \tan \varphi -\tan \gamma \tan \varphi }$

\[=\frac{\sqrt{\frac{xw}{yz}}+\sqrt{\frac{yw}{zx}}+\sqrt{\frac{zw}{xy}}-\sqrt{\frac{w}{1}\times \frac{w}{z}\times \frac{w}{xy}}}{1-\sqrt{\frac{w}{z}\times \frac{w}{z}}-\sqrt{\frac{w}{y}\times \frac{w}{y}}-\sqrt{\frac{yw}{zx}\times \frac{zw}{xy}}}=\frac{\frac{\sqrt{{{x}^{2}}w}+\sqrt{{{y}^{2}}w}+\sqrt{{{z}^{2}}w}}{\sqrt{xyz}}-\sqrt{\frac{{{w}^{3}}}{xyz}}}{1-\sqrt{\frac{{{w}^{2}}}{{{z}^{2}}}}-\sqrt{\frac{{{w}^{2}}}{{{y}^{2}}}}-\sqrt{\frac{{{w}^{2}}}{{{x}^{2}}}}}\]

$=\frac{\frac{x\sqrt{w}+y\sqrt{w}+z\sqrt{w}-w\sqrt{w}}{\sqrt{xyz}}}{1-\frac{w}{z}-\frac{w}{y}-\frac{w}{x}}=\frac{\frac{\sqrt{w}\left( x+y+z-w \right)}{\sqrt{xyz}}}{1-w\left( \frac{1}{x}+\frac{1}{y}+\frac{1}{z} \right)}=\frac{\frac{\sqrt{w}\left( w-w \right)}{\sqrt{xyz}}}{1-w\left( \frac{yz+zx+xy}{xyz} \right)}=0$

So we get $\tan \left( \theta +\gamma +\varphi  \right)=0\Leftrightarrow \tan \left( \theta +\gamma +\varphi  \right)=\tan 0\Leftrightarrow \theta +\gamma +\varphi =0+k\pi $ with $k=1$