Question

If $a,b,c$ be positive real number and the value of $\theta ={\tan }^{-1}\frac{\sqrt{a(a+b+c)}}{bc}+{\tan }^{-1}\frac{\sqrt{b(a+b+c)}}{ca}+{\tan }^{-1}\frac{\sqrt{c(a+b+c)}}{ab}$, $\mathit{t}\mathit{a}\mathit{n}\mathit{\theta }\mathbf{:}$

• A. $0$
• B. $1$
• C. $\frac{a+b+c}{abc}$
• D. $Noneofthese$

Answer 1

The correct option is A

$0$

Finding the value of

Given, $a,b,c$ be positive real number

Let, $x^2=\frac{a+b+c}{abc}$

Now,

$\begin{array}{rcl}\theta & =& {\tan }^{-1}\frac{\sqrt{a(a+b+c)}}{bc}+{\tan }^{-1}\frac{\sqrt{b(a+b+c)}}{ca}+{\tan }^{-1}\frac{\sqrt{c(a+b+c)}}{ab}\\ \theta & =& {\tan }^{-1}\sqrt{a^2{x}^2}+{\tan }^{-1}\sqrt{b^2{x}^2}+{\tan }^{-1}\sqrt{c^2{x}^2}\\ \theta & =& {\tan }^{-1}ax+{\tan }^{-1}bx+{\tan }^{-1}cx\\ \theta & =& {\tan }^{-1}\left[\frac{[ax+bx+cx–abc{x}^3]}{1–x^2(ab+bc+ca)}\right]\\ \tan \theta & =& \frac{x\left[\right(a+b+c)-{\displaystyle \frac{abc(a+b+c)}{abc}}]}{1–x^2(ab+bc+ca)}\mathbf{\{}{\mathit{x}}^{\mathbf{2}}\mathbf{=}\frac{\mathbf{a}\mathbf{+}\mathbf{b}\mathbf{+}\mathbf{c}}{\mathbf{a}\mathbf{b}\mathbf{c}}\mathbf{\}}\\ \tan \theta & =& 0\end{array}$

Hence, correct option is $\mathbf{\left(}\mathit{A}\mathbf{\right)}$