Question

$ta{n}^{-1}\sqrt{\frac{a(a+b+c)}{bc}}+ta{n}^{-1}\sqrt{\frac{b(a+b+c)}{ca}}+ta{n}^{-1}\sqrt{\frac{c(a+b+c)}{ab}}$ is equal to

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

Answer 1

The correct option is D $\pi$

As $\sqrt{\frac{b(a+b+c)}{ca}}\times \sqrt{\frac{c(a+b+c)}{ab}}=\frac{a+b+c}{a}>1$
$ta{n}^{-1}\sqrt{\frac{b(a+b+c)}{ca}}+ta{n}^{-1}\sqrt{\frac{c(a+b+c)}{ab}}=ta{n}^{-1}\left(\frac{\sqrt{\frac{b(a+b+c)}{ca}}+\sqrt{\frac{c(a+b+c)}{ab}}}{1-\sqrt{\frac{b(a+b+c)}{ca}}\times \sqrt{\frac{c(a+b+c)}{ab}}}\right)+\pi =-ta{n}^{-1}\sqrt{\frac{a(a+b+c)}{bc}}+\pi =0\Rightarrow ta{n}^{-1}\sqrt{\frac{a(a+b+c)}{bc}}+ta{n}^{-1}\sqrt{\frac{b(a+b+c)}{ca}}+ta{n}^{-1}\sqrt{\frac{c(a+b+c)}{ab}}=\pi$