Question

If $a,b,c$ are positive real numbers $\theta ={\tan }^{-1}\sqrt{\frac{a(a+b+c)}{bc}}+{\tan }^{-1}\sqrt{\frac{b(a+b+c)}{ca}}+{\tan }^{-1}\sqrt{\frac{c(a+b+c)}{ab}},$ then $\tan \theta$ equals

• A. 0.00
• B. 0
• C. 0.0
• D. 00

Answer 1

Answer: (A), (B), (C), (D)

$\theta ={\tan }^{-1}\sqrt{\frac{a(a+b+c)}{bc}}+{\tan }^{-1}\sqrt{\frac{b(a+b+c)}{ca}}+{\tan }^{-1}\sqrt{\frac{c(a+b+c)}{ab}}$
by using ${\tan }^{-1}x+{\tan }^{-1}y+{\tan }^{-1}z={\tan }^{-1}\left(\frac{x+y+z-xyz}{1-(xy+yz+zx)}\right)$
$\Rightarrow \theta ={\tan }^{-1}\left[\frac{\left(\sqrt{a+b+c}\right)(\sqrt{\frac{a}{bc}}+\sqrt{\frac{b}{ca}}+\sqrt{\frac{c}{ab}})-(a+b+c)\sqrt{\frac{a+b+c}{abc}}}{1-(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})}\right]$
$={\tan }^{-1}\left[\frac{\sqrt{\frac{a+b+c}{abc}}(a+b+c)-(a+b+c)\sqrt{\frac{a+b+c}{abc}}}{1-\frac{(a+b+c)(ab+bc+ca)}{abc}}\right]$
$={\tan }^{-1}\left[\frac{0}{1-\frac{(a+b+c)(ab+bc+ca)}{abc}}\right]$
$\theta ={\tan }^{-1}0$
$\Rightarrow \tan \theta =0$