Question

The value of ${\tan }^{-1}\sqrt{\frac{a(a+b+c)}{bc}}+{\tan }^{-1}\sqrt{\frac{b(a+b+c)}{ca}}+{\tan }^{-1}\sqrt{\frac{c(a+b+c)}{ab}}$ where $a,b,c>0$ is

• A. $\frac{\pi }{4}$
• B. $\frac{\pi }{2}$
• C. $\pi$
• D. $0$

Answer 1

The correct option is C $\pi$

Let $S={\tan }^{-1}\sqrt{\frac{a(a+b+c)}{bc}}+{\tan }^{-1}\sqrt{\frac{b(a+b+c)}{ca}}+{\tan }^{-1}\sqrt{\frac{c(a+b+c)}{ab}}$
Let $x=\sqrt{\frac{a(a+b+c)}{bc}},$ $y=\sqrt{\frac{b(a+b+c)}{ca}}$ and $z=\sqrt{\frac{c(a+b+c)}{ab}}$

$x+y+z=\sqrt{a+b+c}(\sqrt{\frac{a}{bc}}+\sqrt{\frac{b}{ca}}+\sqrt{\frac{c}{ab}})$
$=\sqrt{a+b+c}\left(\frac{a+b+c}{\sqrt{abc}}\right)=\frac{(a+b+c{)}^{3/2}}{(abc{)}^{1/2}}$
$xyz=\sqrt{\frac{a(a+b+c)}{bc}}\cdot \sqrt{\frac{b(a+b+c)}{ca}}\cdot \sqrt{\frac{c(a+b+c)}{ab}}=\frac{(a+b+c{)}^{3/2}}{(abc{)}^{1/2}}$
$\therefore x+y+z=xyz\Rightarrow S=\pi$