Question

If a, b, c be positive real numbers and the value of

$\theta = \tan^{- 1}\sqrt{\frac{a(a + b + c)}{bc}} + \tan^{- 1}\sqrt{\frac{b(a + b + c)}{ca}} + \tan^{- 1}\sqrt{\frac{c(a + b + c)}{ab}}$, then $\tan\theta$ is

• A. 0
• B. 1
• C. $a + b + c$
• D. None of these

Answer 1

Answer: (A)

0

Answer 2

$\theta = \tan^{- 1}\sqrt{\frac{a(a + b + c)}{bc}} + \tan^{- 1}\sqrt{\frac{b(a + b + c)}{ca}} + \tan^{- 1}\sqrt{\frac{c(a + b + c)}{ab}}$

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

∴ $\theta = \tan^{- 1}\sqrt{a^{2}s^{2}} + \tan^{- 1}\sqrt{b^{2}s^{2}} + \tan^{- 1}\sqrt{c^{2}s^{2}}$

$\Rightarrow \theta = \tan^{- 1}(as) + \tan^{- 1}(bs) + \tan^{- 1}(cs)$ ⇒ $\theta = \tan^{- 1}\left\lbrack \frac{as + bs + cs - abcs^{3}}{1 - abs^{2} - bcs^{2} - cas^{2}} \right\rbrack$ $\Rightarrow$ $\tan\theta = s\left\lbrack \frac{(a + b + c) - abcs^{2}}{1 - (ab + bc + ca)s^{2}} \right\rbrack = 0\lbrack\because abcs^{2} = (a + b + c)\rbrack$

Trick : Since it is an identity so it will be true for any value of a,b,c. Let $a = b = c = 1$ then

$\theta = \tan ^ { - 1 } \sqrt { 3 } + \tan ^ { - 1 } \sqrt { 3 } + \tan ^ { - 1 } \sqrt { 3 } = \pi$ $\tan\theta = 0$