Question

If a, b, c are real positive numbers and $\theta ={\tan }^{-1}\sqrt{x}+{\tan }^{-1}\sqrt{y}+{\tan }^{-1}\sqrt{z}$, where $x=\frac{a^2(a+b+c)}{abc},y=\frac{b^2(a+b+c)}{abc},z=\frac{c^2(a+b+c)}{abc}$, prove that $\tan \theta =0.$

Answer 1

We know that ${\tan }^{-1}\left(x\right)+{\tan }^{-1}\left(y\right)+{\tan }^{-1}\left(z\right)={\tan }^{-1}\left(\frac{x+y+z-xyz}{1-xy-yz-zx}\right)$

Given $\theta ={\tan }^{-1}\left(\sqrt{x}\right)+{\tan }^{-1}\left(\sqrt{y}\right)+{\tan }^{-1}\left(\sqrt{z}\right)$

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

$\Rightarrow \tan \theta =\frac{\sqrt{x}+\sqrt{y}+\sqrt{z}-\sqrt{xyz}}{1-\sqrt{xy}-\sqrt{yz}-\sqrt{zx}}$

substituting for $x=\frac{a^2(a+b+c)}{abc}$, $y=\frac{b^2(a+b+c)}{abc}$ and $z=\frac{c^2(a+b+c)}{abc}$ above we get

$\Rightarrow \tan \theta =\frac{\frac{a^2(a+b+c)}{abc}+\frac{b^2(a+b+c)}{abc}+\frac{c^2(a+b+c)}{abc}-\frac{a^2(a+b+c)}{abc}\frac{b^2(a+b+c)}{abc}\frac{c^2(a+b+c)}{abc}}{1-\frac{a^2(a+b+c)}{abc}\frac{b^2(a+b+c)}{abc}-\frac{b^2(a+b+c)}{abc}\frac{c^2(a+b+c)}{abc}-\frac{c^2(a+b+c)}{abc}\frac{a^2(a+b+c)}{abc}}$

$\therefore tan\theta =0$