Question

The value of $\tan ^{-1}\left(\sqrt{\frac{a \lambda}{b c}}\right)+\tan ^{-1}\left(\sqrt{\frac{b \lambda}{c a}}\right)+\tan ^{-1}\left(\sqrt{\frac{c \lambda}{a b}}\right)$, where $a, b, c \in R^{+}$and $\lambda=a+b+c$

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

Answer 1

Answer: (B)

$\left(\sqrt{\frac{a \lambda}{b c}}\right)\left(\sqrt{\frac{b \lambda}{c a}}\right)=\frac{\lambda}{c}=\frac{a+b+c}{c}=1+\frac{b}{c}+\frac{a}{c}>1$
$\Rightarrow \left[\tan ^{-1}\left(\sqrt{\frac{a \lambda}{b c}}\right)+\tan ^{-1}\left(\sqrt{\frac{b \lambda}{c a}}\right)\right]+\tan ^{-1}\left(\sqrt{\frac{c \lambda}{a b}}\right)$
$=\pi+\tan ^{-1} \frac{\sqrt{\frac{a \lambda}{b c}}+\sqrt{\frac{b \lambda}{c a}}}{1-\sqrt{\frac{a \lambda}{b c}}+\sqrt{\frac{b \lambda}{c a}}}+\tan ^{-1}\left(\sqrt{\frac{c \lambda}{a b}}\right)$
$=\pi+\tan ^{-1} \frac{\sqrt{\frac{\lambda}{c}\left[\frac{a+b}{\sqrt{a b}}\right]}}{1-\frac{\lambda}{c}}+\tan ^{-1}\left(\sqrt{\frac{c \lambda}{a b}}\right)$
$=\pi+\tan ^{-1} \frac{\sqrt{\frac{\lambda c}{a b}}[a+b]}{c-\lambda}+\tan ^{-1}\left(\sqrt{\frac{c \lambda}{a b}}\right)$
$=\pi+\tan ^{-1} \frac{\sqrt{\frac{\lambda c}{a b}}[\lambda-c]}{c-\lambda}+\tan ^{-1}\left(\sqrt{\frac{c \lambda}{a b}}\right)$
$=\pi+\tan ^{-1}\left(-\sqrt{\frac{\lambda c}{a b}}\right)+\tan ^{-1}\left(\sqrt{\frac{c \lambda}{a b}}\right)=\pi$