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

If a, b, c be positive real numbers and θ=tan⁡−1akcb+tan⁡−1bkca+tan⁡−1ckab\theta ={{\tan }^{-1}}\sqrt{\dfrac{ak}{cb}}+{{\tan }^{-1}}\sqrt{\dfrac{bk}{ca}}+{{\tan }^{-1}}\sqrt{\dfrac{ck}{ab}}, where k=a+b+ck=a+b+c, then θ\theta equals
A. π2\dfrac{\pi }{2}
B. π4\dfrac{\pi }{4}
C. π\pi
D. none of these

Explanation

Solution

We will solve the given question by considering the terms, akcb\sqrt{\dfrac{ak}{cb}} , tan⁡−1bkca{{\tan }^{-1}}\sqrt{\dfrac{bk}{ca}} , ckab\sqrt{\dfrac{ck}{ab}} and substituting the value of k=a+b+ck=a+b+c in each of them. Then, we will then use the trigonometric identity, tan⁡−1x+tan⁡−1y+tan⁡−1z=tan⁡−1(x+y+z−xyz1−xy−yz−zx){{\tan }^{-1}}x+{{\tan }^{-1}}y+{{\tan }^{-1}}z={{\tan }^{-1}}\left( \dfrac{x+y+z-xyz}{1-xy-yz-zx} \right) . After substituting the values of x as a(a+b+c)cb\sqrt{\dfrac{a\left( a+b+c \right)}{cb}}, y as b(a+b+c)ca\sqrt{\dfrac{b\left( a+b+c \right)}{ca}} and z as c(a+b+c)ab\sqrt{\dfrac{c\left( a+b+c \right)}{ab}} in the above identity and simplifying further, we will be able to get the value of θ\theta .

Complete step by step answer:
In this question, we have been given a, b, c as positive real numbers and an equation,θ=tan⁡−1akcb+tan⁡−1bkca+tan⁡−1ckab\theta ={{\tan }^{-1}}\sqrt{\dfrac{ak}{cb}}+{{\tan }^{-1}}\sqrt{\dfrac{bk}{ca}}+{{\tan }^{-1}}\sqrt{\dfrac{ck}{ab}}, where the values of k=a+b+ck=a+b+c. We have been asked to find the value of θ\theta . So, since the value of k is given as, k=a+b+ck=a+b+c, let us rewrite the given equation by substituting the value of k. So, we have,
θ=tan⁡−1a(a+b+c)cb+tan⁡−1b(a+b+c)ca+tan⁡−1c(a+b+c)ab\theta ={{\tan }^{-1}}\sqrt{\dfrac{a\left( a+b+c \right)}{cb}}+{{\tan }^{-1}}\sqrt{\dfrac{b\left( a+b+c \right)}{ca}}+{{\tan }^{-1}}\sqrt{\dfrac{c\left( a+b+c \right)}{ab}}.
Now, we will use the trigonometric identity of tan⁡−1x+tan⁡−1y+tan⁡−1z=tan⁡−1(x+y+z−xyz1−xy−yz−zx){{\tan }^{-1}}x+{{\tan }^{-1}}y+{{\tan }^{-1}}z={{\tan }^{-1}}\left( \dfrac{x+y+z-xyz}{1-xy-yz-zx} \right) here. So, we will substitute the value of x as a(a+b+c)cb\sqrt{\dfrac{a\left( a+b+c \right)}{cb}}, value of y as b(a+b+c)ca\sqrt{\dfrac{b\left( a+b+c \right)}{ca}} and the value of z as c(a+b+c)ab\sqrt{\dfrac{c\left( a+b+c \right)}{ab}}. So, substituting them in the equation, we get the right hand side or the RHS of the trigonometric identity as,
tan⁡−1a(a+b+c)cb+tan⁡−1b(a+b+c)ca+tan⁡−1c(a+b+c)ab{{\tan }^{-1}}\sqrt{\dfrac{a\left( a+b+c \right)}{cb}}+{{\tan }^{-1}}\sqrt{\dfrac{b\left( a+b+c \right)}{ca}}+{{\tan }^{-1}}\sqrt{\dfrac{c\left( a+b+c \right)}{ab}}
Now, applying the above mentioned trigonometric identity, we will get the left hand side or the LHS as follows,
tan⁡−1(a(a+b+c)cb+b(a+b+c)ca+c(a+b+c)ab−a(a+b+c)cb×b(a+b+c)ca×c(a+b+c)ab1−a(a+b+c)cb×b(a+b+c)ca−b(a+b+c)ca×c(a+b+c)ab−c(a+b+c)ab×a(a+b+c)cb){{\tan }^{-1}}\left( \dfrac{\sqrt{\dfrac{a\left( a+b+c \right)}{cb}}+\sqrt{\dfrac{b\left( a+b+c \right)}{ca}}+\sqrt{\dfrac{c\left( a+b+c \right)}{ab}}-\sqrt{\dfrac{a\left( a+b+c \right)}{cb}}\times \sqrt{\dfrac{b\left( a+b+c \right)}{ca}}\times \sqrt{\dfrac{c\left( a+b+c \right)}{ab}}}{1-\sqrt{\dfrac{a\left( a+b+c \right)}{cb}}\times \sqrt{\dfrac{b\left( a+b+c \right)}{ca}}-\sqrt{\dfrac{b\left( a+b+c \right)}{ca}}\times \sqrt{\dfrac{c\left( a+b+c \right)}{ab}}-\sqrt{\dfrac{c\left( a+b+c \right)}{ab}\times }\sqrt{\dfrac{a\left( a+b+c \right)}{cb}}} \right)
On doing further calculations, we get,
{{\tan }^{-1}}\left\\{ \dfrac{\sqrt{a+b+c}\left( \sqrt{\dfrac{a}{cb}}+\sqrt{\dfrac{b}{ca}}+\sqrt{\dfrac{c}{ab}} \right)-\dfrac{\left( a+b+c \right)\sqrt{\left( a+b+c \right)}}{\sqrt{abc}}}{1-\left( \dfrac{a+b+c}{c} \right)-\left( \dfrac{a+b+c}{a} \right)-\left( \dfrac{a+b+c}{b} \right)} \right\\}
Which can be further written as,
{{\tan }^{-1}}\left\\{ \dfrac{\sqrt{a+b+c}\left( \dfrac{a+b+c}{\sqrt{abc}} \right)-\sqrt{a+b+c}\left( \dfrac{a+b+c}{\sqrt{abc}} \right)}{1-\left( a+b+c \right)\left( \dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c} \right)} \right\\}
On analysing the numerator, we can say that it equals 0, so it can be written as tan⁡−1(0){{\tan }^{-1}}\left( 0 \right).
So, now the equation can be written as,
θ=tan⁡−1(0)\theta ={{\tan }^{-1}}\left( 0 \right)
We will take tan on both the sides and so, we get,
tan⁡θ=0\tan \theta =0
Now, we know that the value of tan⁡π=0\tan \pi =0, so we can say that the value of θ=π\theta =\pi .
Therefore, the correct answer is option C.

Note:
We can also solve this question by using another trigonometric identity, which is tan⁡−1x+tan⁡−1y=tan⁡−1(x+y1−xy){{\tan }^{-1}}x+{{\tan }^{-1}}y={{\tan }^{-1}}\left( \dfrac{x+y}{1-xy} \right). But by using this identity the solution would become tedious, so it is preferable to use the identity given in the solution.