(a) As given 

$\frac{1}{{\tan \theta }} + \tan \theta = m\, $

$\Rightarrow \,1 + {\tan ^2}\theta = m\,\tan \theta $

$ \Rightarrow \,\,{\sec ^2}\theta = m\,\tan \theta $…..$(i) $

and $\sec \theta – \cos \theta = n\,\, \Rightarrow \,\,{\sec ^2}\theta – 1 = n\,\sec \theta $ 

$ \Rightarrow \,\,{\tan ^2}\theta = n\,\,\sec \theta $ 

$ \Rightarrow \,\,{\tan ^4}\theta = {n^2}\,{\sec ^2}\theta = {n^2}.\,m\,\,\tan \theta $      {by $(i)$} 

$ \Rightarrow \,\,{\tan ^3}\theta  = {n^2}m\,,\,\,\,(\,\,\,\tan \theta  \ne 0)$ 

$\Rightarrow \,\,\tan \theta = {({n^2}m)^{1/3}}$…..$(ii)$ 

Also, ${\sec ^2}\theta = m\,\,\tan \theta = m\,{({n^2}m)^{1/3}}$ {by $(i)$ and $(ii)$} 

$\therefore$ Using the identity ${\sec ^2}\theta – {\tan ^2}\theta = 1$ 

$ \Rightarrow \,\,m\,{(m{n^2})^{1/3}} – {({n^2}m)^{2/3}} = 1$ 

$ \Rightarrow \,\,m\,{(m{n^2})^{1/3}} – n\,{(n{m^2})^{1/3}} = 1.$