Question

If ${\tan }^{-1}\sqrt{\left(\frac{a(a+b+c)}{bc}\right)}+{\tan }^{-1}\sqrt{\left(\frac{b(a+b+c)}{ac}\right)}+{\tan }^{-1}\sqrt{\left(\frac{c(a+b+c)}{ab}\right)}=x\pi$ where $a,b,c$ are positive.
Fine the vale of x.

• A. $x=0$
• B. $x=1/2$
• C. $x=1$
• D. $x=2$

Answer 1

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

The correct options are
A $x=0$
C $x=1$
D $x=2$

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