Question

The value of ${\tan }^{-1}a\sqrt{\frac{\lambda }{abc}}+{\tan }^{-1}b\sqrt{\frac{\lambda }{bca}}+{\tan }^{-1}c\sqrt{\frac{\lambda }{cab}}$ where $a,b,c\in R^{+}$ and $\lambda =a+b+c$ is equal to

• A. $\frac{\pi }{4}$
• B. $\pi$
• C. $\frac{\pi }{2}$
• D. None of these

Answer 1

Answer: (B)

$\pi$

${\tan }^{-1}a\sqrt{\frac{\lambda }{abc}}+{\tan }^{-1}b\sqrt{\frac{\lambda }{abc}}+{\tan }^{-1}c\sqrt{\frac{\lambda }{abc}}$
$={\tan }^{-1}a\sqrt{\frac{a+b+c}{abc}}+{\tan }^{-1}b\sqrt{\frac{a+b+c}{abc}}+{\tan }^{-1}c\sqrt{\frac{a+b+c}{abc}}$
Put $\sqrt{\frac{a+b+c}{abc}}=y$
${\tan }^{-1}ay+{\tan }^{-1}by+{\tan }^{-1}cy$
$=\pi -{\tan }^{-1}\left[\frac{ay+by+cy-abc{y}^3}{1-y^2(ab+bc+ca)}\right]$
$=\pi -{\tan }^{-1}\left[\frac{y(a+b+c-abc{y}^2)}{1-y^2(ab+bc+ca)}\right]$
$=\pi -{\tan }^{-1}\left(0\right)$ ($\because a+b+c=abc{y}^2)$
$=\pi$