Question

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

Answer 1

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

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

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