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Question: If a, b and c are positive real numbers and \(\theta ={{\tan }^{-1}}\left( \sqrt{\dfrac{a\left( a+b+...

If a, b and c are positive real numbers and θ=tan⁡−1(a(a+b+c)bc)+tan⁡−1(b(a+b+c)ca)+tan⁡−1(c(a+b+c)ab)\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 θ\theta ?
(a) π4\dfrac{\pi }{4},
(b) π2\dfrac{\pi }{2},
(c) π3\dfrac{\pi }{3},
(d) 0.

Explanation

Solution

We start solving the problem by recalling the sum of inverse of the tangent of 3 values i.e., tan⁡−1(x)+tan⁡−1(y)+tan⁡−1(z)=tan⁡−1(x+y+z−xyz1−xy−yz−zx){{\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 θ\theta . We then make necessary arrangements inside the inverse of tangent and use the facts ab=ab\sqrt{\dfrac{a}{b}}=\dfrac{\sqrt{a}}{\sqrt{b}}, ab=a2ba\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 θ\theta .

Complete step-by-step solution:
According to the problem, we have given a, b and c are positive real numbers and θ=tan⁡−1(a(a+b+c)bc)+tan⁡−1(b(a+b+c)ca)+tan⁡−1(c(a+b+c)ab)\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 θ\theta .
We know that tan⁡−1(x)+tan⁡−1(y)+tan⁡−1(z)=tan⁡−1(x+y+z−xyz1−xy−yz−zx){{\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 θ\theta .
So, we have θ=tan⁡−1(a(a+b+c)bc+b(a+b+c)ca+c(a+b+c)ab−(a(a+b+c)bc×b(a+b+c)ca×c(a+b+c)ab)1−(a(a+b+c)bc×b(a+b+c)ca)−(b(a+b+c)ca×c(a+b+c)ab)−(c(a+b+c)ab×a(a+b+c)bc))\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).
⇒θ=tan⁡−1(a2(a+b+c)abc+b2(a+b+c)abc+c2(a+b+c)abc−(abc(a+b+c)3(abc)2)1−(ab(a+b+c)2abc2)−(bc(a+b+c)2a2bc)−(ac(a+b+c)2ab2c))\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 ab=ab\sqrt{\dfrac{a}{b}}=\dfrac{\sqrt{a}}{\sqrt{b}} to solve for the angle θ\theta .
⇒θ=tan⁡−1(a2(a+b+c)abc+b2(a+b+c)abc+c2(a+b+c)abc−((a+b+c)3(abc))1−(a2b2(a+b+c)2a2b2c2)−(b2c2(a+b+c)2a2b2c2)−(a2c2(a+b+c)2a2b2c2))\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).
⇒θ=tan⁡−1(a(a+b+c)abc+b(a+b+c)abc+c(a+b+c)abc−((a+b+c)3(abc))1−(a2b2(a+b+c)2a2b2c2)−(b2c2(a+b+c)2a2b2c2)−(c2a2(a+b+c)2a2b2c2))\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).
⇒θ=tan⁡−1((((a+b+c)abc)×(a+b+c))−((a+b+c)3(abc))1−(ab(a+b+c)abc)−(bc(a+b+c)abc)−(ca(a+b+c)abc))\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 ab=a2ba\sqrt{b}=\sqrt{{{a}^{2}}b} to solve for the angle θ\theta .
⇒θ=tan⁡−1((((a+b+c)×(a+b+c)2abc))−((a+b+c)3(abc))1−((ab(a+b+c)abc)+(bc(a+b+c)abc)+(ca(a+b+c)abc)))\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).
⇒θ=tan⁡−1(((a+b+c)3abc)−((a+b+c)3(abc))1−((ab(a+b+c)+bc(a+b+c)+ca(a+b+c)abc)))\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 ab=ab\dfrac{\sqrt{a}}{\sqrt{b}}=\sqrt{\dfrac{a}{b}} to solve for the angle θ\theta .
⇒θ=tan⁡−1(((a+b+c)3(abc))−((a+b+c)3(abc))1−((a+b+c)(ab+bc+ca)abc))\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.
⇒θ=tan⁡−1(0(abc−((a+b+c)(ab+bc+ca))abc))\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 0a=0\dfrac{0}{a}=0.
⇒θ=tan⁡−1(0)\Rightarrow \theta ={{\tan }^{-1}}\left( 0 \right).
⇒θ=0o\Rightarrow \theta ={{0}^{o}}.
We have found the value of the angle θ\theta as 0o{{0}^{o}}.
∴\therefore The value of the angles θ\theta is 0o{{0}^{o}}.
The correct option for the given problem is (d).

Note: We can also solve this problem by assuming α=tan⁡−1(a(a+b+c)bc)\alpha ={{\tan }^{-1}}\left( \sqrt{\dfrac{a\left( a+b+c \right)}{bc}} \right), β=tan⁡−1(b(a+b+c)ca)\beta ={{\tan }^{-1}}\left( \sqrt{\dfrac{b\left( a+b+c \right)}{ca}} \right), δ=tan⁡−1(c(a+b+c)ab)\delta ={{\tan }^{-1}}\left( \sqrt{\dfrac{c\left( a+b+c \right)}{ab}} \right) and using the formula tan⁡(α+β+δ)=tan⁡α+tan⁡β+tan⁡δ−tan⁡αtan⁡βtan⁡δ1−tan⁡αtan⁡β−tan⁡βtan⁡δ−tan⁡δtan⁡α\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 00 and π2\dfrac{\pi }{2}, we opted for 00. 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.