If a, b and c are positive real numbers and $\theta ={{\tan }^{-1}}\left( \sqrt{\dfrac{a\left( a+b+c \right)}{bc}} \right)+{{\tan }^{-1}}\left( \sqrt{\dfrac{b\left( a+b+c \right)}{ca}} \right)+{{\tan }^{-1}}\left( \sqrt{\dfrac{c\left( a+b+c \right)}{ab}} \right)$, then find the value of the angle $\theta $?
(a) $\dfrac{\pi }{4}$,
(b) $\dfrac{\pi }{2}$,
(c) $\dfrac{\pi }{3}$,
(d) 0.
Answer
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Hint: We start solving the problem by recalling the sum of inverse of the tangent of 3 values i.e., ${{\tan }^{-1}}\left( x \right)+{{\tan }^{-1}}\left( y \right)+{{\tan }^{-1}}\left( z \right)={{\tan }^{-1}}\left( \dfrac{x+y+z-xyz}{1-xy-yz-zx} \right)$. We apply this for the equation given in the problem to find the angle $\theta $. We then make necessary arrangements inside the inverse of tangent and use the facts $\sqrt{\dfrac{a}{b}}=\dfrac{\sqrt{a}}{\sqrt{b}}$, $a\sqrt{b}=\sqrt{{{a}^{2}}b}$ to get a value in the numerator. We then make necessary calculations to find the value of the required angle $\theta $.
Complete step-by-step solution:
According to the problem, we have given a, b and c are positive real numbers and $\theta ={{\tan }^{-1}}\left( \sqrt{\dfrac{a\left( a+b+c \right)}{bc}} \right)+{{\tan }^{-1}}\left( \sqrt{\dfrac{b\left( a+b+c \right)}{ca}} \right)+{{\tan }^{-1}}\left( \sqrt{\dfrac{c\left( a+b+c \right)}{ab}} \right)$. We need to find the value of angle $\theta $.
We know that ${{\tan }^{-1}}\left( x \right)+{{\tan }^{-1}}\left( y \right)+{{\tan }^{-1}}\left( z \right)={{\tan }^{-1}}\left( \dfrac{x+y+z-xyz}{1-xy-yz-zx} \right)$. We use this to calculate the value of angle $\theta $.
So, we have \[\theta ={{\tan }^{-1}}\left( \dfrac{\sqrt{\dfrac{a\left( a+b+c \right)}{bc}}+\sqrt{\dfrac{b\left( a+b+c \right)}{ca}}+\sqrt{\dfrac{c\left( a+b+c \right)}{ab}}-\left( \sqrt{\dfrac{a\left( a+b+c \right)}{bc}}\times \sqrt{\dfrac{b\left( a+b+c \right)}{ca}}\times \sqrt{\dfrac{c\left( a+b+c \right)}{ab}} \right)}{1-\left( \sqrt{\dfrac{a\left( a+b+c \right)}{bc}}\times \sqrt{\dfrac{b\left( a+b+c \right)}{ca}} \right)-\left( \sqrt{\dfrac{b\left( a+b+c \right)}{ca}}\times \sqrt{\dfrac{c\left( a+b+c \right)}{ab}} \right)-\left( \sqrt{\dfrac{c\left( a+b+c \right)}{ab}}\times \sqrt{\dfrac{a\left( a+b+c \right)}{bc}} \right)} \right)\].
$\Rightarrow \theta ={{\tan }^{-1}}\left( \dfrac{\sqrt{\dfrac{{{a}^{2}}\left( a+b+c \right)}{abc}}+\sqrt{\dfrac{{{b}^{2}}\left( a+b+c \right)}{abc}}+\sqrt{\dfrac{{{c}^{2}}\left( a+b+c \right)}{abc}}-\left( \sqrt{\dfrac{abc{{\left( a+b+c \right)}^{3}}}{{{\left( abc \right)}^{2}}}} \right)}{1-\left( \sqrt{\dfrac{ab{{\left( a+b+c \right)}^{2}}}{ab{{c}^{2}}}} \right)-\left( \sqrt{\dfrac{bc{{\left( a+b+c \right)}^{2}}}{{{a}^{2}}bc}} \right)-\left( \sqrt{\dfrac{ac{{\left( a+b+c \right)}^{2}}}{a{{b}^{2}}c}} \right)} \right)$.
We use the fact that $\sqrt{\dfrac{a}{b}}=\dfrac{\sqrt{a}}{\sqrt{b}}$ to solve for the angle $\theta $.
$\Rightarrow \theta ={{\tan }^{-1}}\left( \dfrac{\dfrac{\sqrt{{{a}^{2}}\left( a+b+c \right)}}{\sqrt{abc}}+\dfrac{\sqrt{{{b}^{2}}\left( a+b+c \right)}}{\sqrt{abc}}+\dfrac{\sqrt{{{c}^{2}}\left( a+b+c \right)}}{\sqrt{abc}}-\left( \sqrt{\dfrac{{{\left( a+b+c \right)}^{3}}}{\left( abc \right)}} \right)}{1-\left( \sqrt{\dfrac{{{a}^{2}}{{b}^{2}}{{\left( a+b+c \right)}^{2}}}{{{a}^{2}}{{b}^{2}}{{c}^{2}}}} \right)-\left( \sqrt{\dfrac{{{b}^{2}}{{c}^{2}}{{\left( a+b+c \right)}^{2}}}{{{a}^{2}}{{b}^{2}}{{c}^{2}}}} \right)-\left( \sqrt{\dfrac{{{a}^{2}}{{c}^{2}}{{\left( a+b+c \right)}^{2}}}{{{a}^{2}}{{b}^{2}}{{c}^{2}}}} \right)} \right)$.
$\Rightarrow \theta ={{\tan }^{-1}}\left( \dfrac{\dfrac{a\sqrt{\left( a+b+c \right)}}{\sqrt{abc}}+\dfrac{b\sqrt{\left( a+b+c \right)}}{\sqrt{abc}}+\dfrac{c\sqrt{\left( a+b+c \right)}}{\sqrt{abc}}-\left( \sqrt{\dfrac{{{\left( a+b+c \right)}^{3}}}{\left( abc \right)}} \right)}{1-\left( \dfrac{\sqrt{{{a}^{2}}{{b}^{2}}{{\left( a+b+c \right)}^{2}}}}{\sqrt{{{a}^{2}}{{b}^{2}}{{c}^{2}}}} \right)-\left( \dfrac{\sqrt{{{b}^{2}}{{c}^{2}}{{\left( a+b+c \right)}^{2}}}}{\sqrt{{{a}^{2}}{{b}^{2}}{{c}^{2}}}} \right)-\left( \dfrac{\sqrt{{{c}^{2}}{{a}^{2}}{{\left( a+b+c \right)}^{2}}}}{\sqrt{{{a}^{2}}{{b}^{2}}{{c}^{2}}}} \right)} \right)$.
$\Rightarrow \theta ={{\tan }^{-1}}\left( \dfrac{\left( \left( \dfrac{\sqrt{\left( a+b+c \right)}}{\sqrt{abc}} \right)\times \left( a+b+c \right) \right)-\left( \sqrt{\dfrac{{{\left( a+b+c \right)}^{3}}}{\left( abc \right)}} \right)}{1-\left( \dfrac{ab\left( a+b+c \right)}{abc} \right)-\left( \dfrac{bc\left( a+b+c \right)}{abc} \right)-\left( \dfrac{ca\left( a+b+c \right)}{abc} \right)} \right)$.
We use the fact that $a\sqrt{b}=\sqrt{{{a}^{2}}b}$ to solve for the angle $\theta $.
$\Rightarrow \theta ={{\tan }^{-1}}\left( \dfrac{\left( \left( \dfrac{\sqrt{\left( a+b+c \right)\times {{\left( a+b+c \right)}^{2}}}}{\sqrt{abc}} \right) \right)-\left( \sqrt{\dfrac{{{\left( a+b+c \right)}^{3}}}{\left( abc \right)}} \right)}{1-\left( \left( \dfrac{ab\left( a+b+c \right)}{abc} \right)+\left( \dfrac{bc\left( a+b+c \right)}{abc} \right)+\left( \dfrac{ca\left( a+b+c \right)}{abc} \right) \right)} \right)$.
$\Rightarrow \theta ={{\tan }^{-1}}\left( \dfrac{\left( \dfrac{\sqrt{{{\left( a+b+c \right)}^{3}}}}{\sqrt{abc}} \right)-\left( \sqrt{\dfrac{{{\left( a+b+c \right)}^{3}}}{\left( abc \right)}} \right)}{1-\left( \left( \dfrac{ab\left( a+b+c \right)+bc\left( a+b+c \right)+ca\left( a+b+c \right)}{abc} \right) \right)} \right)$.
We use the fact that $\dfrac{\sqrt{a}}{\sqrt{b}}=\sqrt{\dfrac{a}{b}}$ to solve for the angle $\theta $.
\[\Rightarrow \theta ={{\tan }^{-1}}\left( \dfrac{\left( \sqrt{\dfrac{{{\left( a+b+c \right)}^{3}}}{\left( abc \right)}} \right)-\left( \sqrt{\dfrac{{{\left( a+b+c \right)}^{3}}}{\left( abc \right)}} \right)}{1-\left( \dfrac{\left( a+b+c \right)\left( ab+bc+ca \right)}{abc} \right)} \right)\].
We can see that the two terms in the numerator are the same.
\[\Rightarrow \theta ={{\tan }^{-1}}\left( \dfrac{0}{\left( \dfrac{abc-\left( \left( a+b+c \right)\left( ab+bc+ca \right) \right)}{abc} \right)} \right)\].
We know that $\dfrac{0}{a}=0$.
\[\Rightarrow \theta ={{\tan }^{-1}}\left( 0 \right)\].
\[\Rightarrow \theta ={{0}^{o}}\].
We have found the value of the angle $\theta $ as ${{0}^{o}}$.
$\therefore$ The value of the angles $\theta $ is ${{0}^{o}}$.
The correct option for the given problem is (d).
Note: We can also solve this problem by assuming $\alpha ={{\tan }^{-1}}\left( \sqrt{\dfrac{a\left( a+b+c \right)}{bc}} \right)$, $\beta ={{\tan }^{-1}}\left( \sqrt{\dfrac{b\left( a+b+c \right)}{ca}} \right)$, $\delta ={{\tan }^{-1}}\left( \sqrt{\dfrac{c\left( a+b+c \right)}{ab}} \right)$ and using the formula $\tan \left( \alpha +\beta +\delta \right)=\dfrac{\tan \alpha +\tan \beta +\tan \delta -\tan \alpha \tan \beta \tan \delta }{1-\tan \alpha \tan \beta -\tan \beta \tan \delta -\tan \delta \tan \alpha }$ to get the required value. Since the options are between $0$ and $\dfrac{\pi }{2}$, we opted for $0$. If the problem is about solving the trigonometric equation then we have to write the general solution that represents all the angles whose tangent is 0.
Complete step-by-step solution:
According to the problem, we have given a, b and c are positive real numbers and $\theta ={{\tan }^{-1}}\left( \sqrt{\dfrac{a\left( a+b+c \right)}{bc}} \right)+{{\tan }^{-1}}\left( \sqrt{\dfrac{b\left( a+b+c \right)}{ca}} \right)+{{\tan }^{-1}}\left( \sqrt{\dfrac{c\left( a+b+c \right)}{ab}} \right)$. We need to find the value of angle $\theta $.
We know that ${{\tan }^{-1}}\left( x \right)+{{\tan }^{-1}}\left( y \right)+{{\tan }^{-1}}\left( z \right)={{\tan }^{-1}}\left( \dfrac{x+y+z-xyz}{1-xy-yz-zx} \right)$. We use this to calculate the value of angle $\theta $.
So, we have \[\theta ={{\tan }^{-1}}\left( \dfrac{\sqrt{\dfrac{a\left( a+b+c \right)}{bc}}+\sqrt{\dfrac{b\left( a+b+c \right)}{ca}}+\sqrt{\dfrac{c\left( a+b+c \right)}{ab}}-\left( \sqrt{\dfrac{a\left( a+b+c \right)}{bc}}\times \sqrt{\dfrac{b\left( a+b+c \right)}{ca}}\times \sqrt{\dfrac{c\left( a+b+c \right)}{ab}} \right)}{1-\left( \sqrt{\dfrac{a\left( a+b+c \right)}{bc}}\times \sqrt{\dfrac{b\left( a+b+c \right)}{ca}} \right)-\left( \sqrt{\dfrac{b\left( a+b+c \right)}{ca}}\times \sqrt{\dfrac{c\left( a+b+c \right)}{ab}} \right)-\left( \sqrt{\dfrac{c\left( a+b+c \right)}{ab}}\times \sqrt{\dfrac{a\left( a+b+c \right)}{bc}} \right)} \right)\].
$\Rightarrow \theta ={{\tan }^{-1}}\left( \dfrac{\sqrt{\dfrac{{{a}^{2}}\left( a+b+c \right)}{abc}}+\sqrt{\dfrac{{{b}^{2}}\left( a+b+c \right)}{abc}}+\sqrt{\dfrac{{{c}^{2}}\left( a+b+c \right)}{abc}}-\left( \sqrt{\dfrac{abc{{\left( a+b+c \right)}^{3}}}{{{\left( abc \right)}^{2}}}} \right)}{1-\left( \sqrt{\dfrac{ab{{\left( a+b+c \right)}^{2}}}{ab{{c}^{2}}}} \right)-\left( \sqrt{\dfrac{bc{{\left( a+b+c \right)}^{2}}}{{{a}^{2}}bc}} \right)-\left( \sqrt{\dfrac{ac{{\left( a+b+c \right)}^{2}}}{a{{b}^{2}}c}} \right)} \right)$.
We use the fact that $\sqrt{\dfrac{a}{b}}=\dfrac{\sqrt{a}}{\sqrt{b}}$ to solve for the angle $\theta $.
$\Rightarrow \theta ={{\tan }^{-1}}\left( \dfrac{\dfrac{\sqrt{{{a}^{2}}\left( a+b+c \right)}}{\sqrt{abc}}+\dfrac{\sqrt{{{b}^{2}}\left( a+b+c \right)}}{\sqrt{abc}}+\dfrac{\sqrt{{{c}^{2}}\left( a+b+c \right)}}{\sqrt{abc}}-\left( \sqrt{\dfrac{{{\left( a+b+c \right)}^{3}}}{\left( abc \right)}} \right)}{1-\left( \sqrt{\dfrac{{{a}^{2}}{{b}^{2}}{{\left( a+b+c \right)}^{2}}}{{{a}^{2}}{{b}^{2}}{{c}^{2}}}} \right)-\left( \sqrt{\dfrac{{{b}^{2}}{{c}^{2}}{{\left( a+b+c \right)}^{2}}}{{{a}^{2}}{{b}^{2}}{{c}^{2}}}} \right)-\left( \sqrt{\dfrac{{{a}^{2}}{{c}^{2}}{{\left( a+b+c \right)}^{2}}}{{{a}^{2}}{{b}^{2}}{{c}^{2}}}} \right)} \right)$.
$\Rightarrow \theta ={{\tan }^{-1}}\left( \dfrac{\dfrac{a\sqrt{\left( a+b+c \right)}}{\sqrt{abc}}+\dfrac{b\sqrt{\left( a+b+c \right)}}{\sqrt{abc}}+\dfrac{c\sqrt{\left( a+b+c \right)}}{\sqrt{abc}}-\left( \sqrt{\dfrac{{{\left( a+b+c \right)}^{3}}}{\left( abc \right)}} \right)}{1-\left( \dfrac{\sqrt{{{a}^{2}}{{b}^{2}}{{\left( a+b+c \right)}^{2}}}}{\sqrt{{{a}^{2}}{{b}^{2}}{{c}^{2}}}} \right)-\left( \dfrac{\sqrt{{{b}^{2}}{{c}^{2}}{{\left( a+b+c \right)}^{2}}}}{\sqrt{{{a}^{2}}{{b}^{2}}{{c}^{2}}}} \right)-\left( \dfrac{\sqrt{{{c}^{2}}{{a}^{2}}{{\left( a+b+c \right)}^{2}}}}{\sqrt{{{a}^{2}}{{b}^{2}}{{c}^{2}}}} \right)} \right)$.
$\Rightarrow \theta ={{\tan }^{-1}}\left( \dfrac{\left( \left( \dfrac{\sqrt{\left( a+b+c \right)}}{\sqrt{abc}} \right)\times \left( a+b+c \right) \right)-\left( \sqrt{\dfrac{{{\left( a+b+c \right)}^{3}}}{\left( abc \right)}} \right)}{1-\left( \dfrac{ab\left( a+b+c \right)}{abc} \right)-\left( \dfrac{bc\left( a+b+c \right)}{abc} \right)-\left( \dfrac{ca\left( a+b+c \right)}{abc} \right)} \right)$.
We use the fact that $a\sqrt{b}=\sqrt{{{a}^{2}}b}$ to solve for the angle $\theta $.
$\Rightarrow \theta ={{\tan }^{-1}}\left( \dfrac{\left( \left( \dfrac{\sqrt{\left( a+b+c \right)\times {{\left( a+b+c \right)}^{2}}}}{\sqrt{abc}} \right) \right)-\left( \sqrt{\dfrac{{{\left( a+b+c \right)}^{3}}}{\left( abc \right)}} \right)}{1-\left( \left( \dfrac{ab\left( a+b+c \right)}{abc} \right)+\left( \dfrac{bc\left( a+b+c \right)}{abc} \right)+\left( \dfrac{ca\left( a+b+c \right)}{abc} \right) \right)} \right)$.
$\Rightarrow \theta ={{\tan }^{-1}}\left( \dfrac{\left( \dfrac{\sqrt{{{\left( a+b+c \right)}^{3}}}}{\sqrt{abc}} \right)-\left( \sqrt{\dfrac{{{\left( a+b+c \right)}^{3}}}{\left( abc \right)}} \right)}{1-\left( \left( \dfrac{ab\left( a+b+c \right)+bc\left( a+b+c \right)+ca\left( a+b+c \right)}{abc} \right) \right)} \right)$.
We use the fact that $\dfrac{\sqrt{a}}{\sqrt{b}}=\sqrt{\dfrac{a}{b}}$ to solve for the angle $\theta $.
\[\Rightarrow \theta ={{\tan }^{-1}}\left( \dfrac{\left( \sqrt{\dfrac{{{\left( a+b+c \right)}^{3}}}{\left( abc \right)}} \right)-\left( \sqrt{\dfrac{{{\left( a+b+c \right)}^{3}}}{\left( abc \right)}} \right)}{1-\left( \dfrac{\left( a+b+c \right)\left( ab+bc+ca \right)}{abc} \right)} \right)\].
We can see that the two terms in the numerator are the same.
\[\Rightarrow \theta ={{\tan }^{-1}}\left( \dfrac{0}{\left( \dfrac{abc-\left( \left( a+b+c \right)\left( ab+bc+ca \right) \right)}{abc} \right)} \right)\].
We know that $\dfrac{0}{a}=0$.
\[\Rightarrow \theta ={{\tan }^{-1}}\left( 0 \right)\].
\[\Rightarrow \theta ={{0}^{o}}\].
We have found the value of the angle $\theta $ as ${{0}^{o}}$.
$\therefore$ The value of the angles $\theta $ is ${{0}^{o}}$.
The correct option for the given problem is (d).
Note: We can also solve this problem by assuming $\alpha ={{\tan }^{-1}}\left( \sqrt{\dfrac{a\left( a+b+c \right)}{bc}} \right)$, $\beta ={{\tan }^{-1}}\left( \sqrt{\dfrac{b\left( a+b+c \right)}{ca}} \right)$, $\delta ={{\tan }^{-1}}\left( \sqrt{\dfrac{c\left( a+b+c \right)}{ab}} \right)$ and using the formula $\tan \left( \alpha +\beta +\delta \right)=\dfrac{\tan \alpha +\tan \beta +\tan \delta -\tan \alpha \tan \beta \tan \delta }{1-\tan \alpha \tan \beta -\tan \beta \tan \delta -\tan \delta \tan \alpha }$ to get the required value. Since the options are between $0$ and $\dfrac{\pi }{2}$, we opted for $0$. If the problem is about solving the trigonometric equation then we have to write the general solution that represents all the angles whose tangent is 0.
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