Explanation

(a)
Given that$,\ \frac{1}{\tan\theta} + \tan{\theta = m}$
$\Rightarrow 1 + \tan^{2}{\theta = m\tan\theta}$
$\Rightarrow \sec^{2}{\theta = m\tan\theta}$ …(i)
and${\ sec}{\theta - \cos{\theta = n}}$
$\Rightarrow \sec^{2}{\theta - 1 = n\sec\theta}$
$\Rightarrow \tan^{2}{\theta = n\sec\theta}$
$\Rightarrow \tan^{4}{\theta = n^{2}\sec^{2}{\theta = n^{2} \bullet m\tan{\theta\ \ }}}$[from Eq.(i)]
$\Rightarrow \tan^{3}{\theta = n^{2}m\ \ \ (\because\tan{\theta \neq 0)}}$
$\Rightarrow {tan\ }{\theta = \left( n^{2}m \right)^{1/3}}$ …(ii)
From Eq. (i), we get
$\sec^{2}{\theta = m\ \left( n^{2}m \right)^{1/3}}$
As we know that, $\sec^{2}{\theta - \tan^{2}{\theta = 1}}$
$\Rightarrow m\left( mn^{2} \right)^{1/3} - \ \left( n^{2}m \right)^{2/3} = 1$
$\Rightarrow m\left( mn^{2} \right)^{1/3} - \ n\left( nm^{2} \right)^{1/3} = 1$