Question

The value of ${\tan }^{-1}\sqrt{\frac{a(a+b+c)}{bc}}+{\tan }^{-1}\sqrt{\frac{b(a+b+c)}{ac}}+{\tan }^{-1}\sqrt{\frac{c(a+b+c)}{ab}}$ is equal to

• A. $\pi /4$
• B. $\pi /2$
• C. $\pi$
• D. $0$

Answer 1

The correct option is C $\pi$

Let $k=\sqrt{\frac{a+b+c}{abc}}⟹k^2=\frac{a+b+c}{abc}⟹abc{k}^2=a+b+c$

$\theta ={\text{tan}}^{-1}\sqrt{\frac{a(a+b+c)}{bc}}+{\text{tan}}^{-1}\sqrt{\frac{b(a+b+c)}{ca}}+{\text{tan}}^{-1}\sqrt{\frac{c(a+b+c)}{ab}}$

$={\text{tan}}^{-1}\sqrt{\frac{a^2(a+b+c)}{abc}}+{\text{tan}}^{-1}\sqrt{\frac{b^2(a+b+c)}{abc}}+{\text{tan}}^{-1}\sqrt{\frac{c^2(a+b+c)}{abc}}$

$={\text{tan}}^{-1}ak+{\text{tan}}^{-1}bk+{\text{tan}}^{-1}ck$

$={\text{tan}}^{-1}(\frac{k(a+b+c)-abc{k}^3}{1-(ab+bc+ca)k^2})$

$={\text{tan}}^{-1}(\frac{k\left(\right(a+b+c)-abc{k}^2)}{1-(ab+bc+ca)k^2})$

$={\text{tan}}^{-1}(\frac{k\left(\right(a+b+c)-(a+b+c\left)\right)}{1-(ab+bc+ca)k^2})$

$={\text{tan}}^{-1}\left(0\right)$