Question

Let $a,b,c$ be a positive real numbers $\theta ={\tan }^{-1}\sqrt{\frac{a(a+b+c)}{bc}}+{\tan }^{-1}\sqrt{\frac{b(a+b+c)}{ca}}+{\tan }^{-1}\sqrt{\frac{c(a+b+c)}{ab}}$, then $\tan \theta$

• A. $0$
• B. $3\pi$
• C. $1$
• D. $4\pi$

Answer 1

Answer: (A)

$0$

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

$\alpha ={\tan }^{-1}\sqrt{\frac{a(a+b+c)}{bc}}$

$\beta ={\tan }^{-1}\sqrt{\frac{b(a+b+c)}{ac}}$

$\gamma ={\tan }^{-1}\sqrt{\frac{c(a+b+c)}{ab}}$

$\Rightarrow \tan \alpha =\sqrt{\frac{a(a+b+c)}{bc}}$

$\Rightarrow \tan \beta =\sqrt{\frac{b(a+b+c)}{ac}}$

$\Rightarrow \tan \gamma =\sqrt{\frac{c(a+b+c)}{ab}}$

$\tan \alpha +\tan \beta +\tan \gamma =\sqrt{\frac{a(a+b+c)}{bc}}+\sqrt{\frac{b(a+b+c)}{ac}}+\sqrt{\frac{c(a+b+c)}{ab}}$

$=\frac{{(a+b+c)}^{\frac{3}{2}}}{\sqrt{abc}}$

$=\tan \alpha \tan \beta \tan \gamma$

$\Rightarrow \tan \theta =\left[\frac{(\tan \alpha +\tan \beta +\tan \gamma )-\tan \alpha \tan \beta \tan \gamma }{1-\tan \alpha \tan \beta -\tan \beta \tan \gamma -\tan \gamma \tan \alpha }\right]$

$\Rightarrow \tan \theta =0$