Explanation

(c)
Now, $\tan^{- 1}{\left( \frac{1}{2} \right) + \tan^{- 1}\left( \frac{1}{3} \right) + \tan^{- 1}\frac{7}{8}}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \tan^{- 1}\left( \frac{\frac{1}{2} + \frac{1}{3}}{1 - \frac{1}{2}.\frac{1}{3}} \right) + \tan^{- 1}\frac{7}{8}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \tan^{- 1}\left( \frac{\frac{5}{6}}{\frac{5}{6}} \right) + {tan\ }^{- 1}\frac{7}{8}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ = \tan^{- 1}{(1) + \tan^{- 1}\frac{7}{8}}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ = \tan^{- 1}\left( \frac{1 + \frac{7}{8}}{1 - \frac{7}{8}} \right)$
$\ \ \ \ \ \ \ \ \ \ \ \ = \tan^{- 1}{(15)}$