Explanation

(a)
$\theta = \tan^{- 1}{\sqrt{\frac{a(a + b + c)}{bc}} + \tan^{- 1}{\sqrt{\frac{b(a + b + c)}{ca}} + \tan^{- 1}\sqrt{\frac{c(a + b + c)}{ab}}}}$
Let $s^{2} = \frac{a + b + c}{abc}$
Hence, $\theta = \tan^{- 1}{\sqrt{a^{2}s^{2}} + \tan^{- 1}{\sqrt{b^{2}s^{2}} + \tan^{- 1}\sqrt{c^{2}s^{2}}}}$
$= \tan^{- 1}{(as) + \tan^{- 1}{(bs) + \tan^{- 1}{(cs)}}}$
$= \tan^{- 1}\left\lbrack \frac{as + bs + cs - abcs^{3}}{1 - abs^{2} - acs^{2} - bcs^{2}} \right\rbrack$
Hence, $\tan{\theta = \left\lbrack \frac{s\lbrack a + b + c) - abcs^{2}\rbrack}{1 - (ab + bc + ca)s^{2}} \right\rbrack}$
$= \left\lbrack \frac{s\lbrack(a + b + c) - (a + b + c)\rbrack}{1 - s^{2}(ab + bc + ca)} \right\rbrack = 0$