Question

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

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

Answer 1

Answer: (C)

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