Question

The value of ${\tan }^{-1}\left(\sqrt{\frac{a\lambda }{bc}}\right)+{\tan }^{-1}\left(\sqrt{\frac{b\lambda }{ca}}\right)+{\tan }^{-1}\left(\sqrt{\frac{c\lambda }{ab}}\right)$, 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)

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