दिये गये अनुसार,

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

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

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

व $\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 $    {$(i)$ द्वारा}

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

तथा ${\sec ^2}\theta  = m\,\,\tan \theta  = m\,{({n^2}m)^{1/3}}$     {$(i)$ व $(ii)$ से}

$\therefore$ सर्वसमिका ${\sec ^2}\theta  – {\tan ^2}\theta  = 1$ से,

$ \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.$