Question

If $a,b,c$ are positive , then $\sum ta{n}^{-1}\sqrt{\frac{a(a+b+c)}{bc}}=$

• A. $\pi$
• B. $\frac{3\pi }{2}$
• C. $\frac{3\pi }{4}$
• D. $3$

Answer 1

The correct option is A $\pi$

Let $A=\sqrt{\frac{a(a+b+c)}{bc}}$
$ABC=\frac{a+b+c}{abc}$
$A+B+C=\sqrt{\frac{a+b+c}{abc}}(a+b+c)$
$AB=\frac{a+b+c}{c}$
$BC=\frac{a+b+c}{a}$
$\to ta{n}^{-1}A+ta{n}^{-1}B+ta{n}^{-1}C$
$=ta{n}^{-1}\left(\frac{A+B}{1-AB}\right)+ta{n}^{-1}C$
$=ta{n}^{-1}\left(\frac{A+B}{1-AB}\right)+ta{n}^{-1}C$
$ta{n}^{-1}\frac{(\frac{A+B}{1-AB}+C)}{1-\left(\frac{AC+BC}{1-AB}\right)}$
$=ta{n}^{-1}\left(\frac{A+B+C-ABC}{1-AB-BC-AC}\right)$
$=ta{n}^{-1}\frac{(\frac{a+b+c}{abc}-(a+b+c\left)\sqrt{\frac{a+b+c}{abc}}\right)}{1-(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})}$
$=ta{n}^{-1}\left(0\right)$
$=\pi$