Correct Answer: $2m$

Solution : We know that $\cos\theta+\sin\theta=m$ and $\sec\theta+\operatorname{cosec \theta}=n$.
$⇒n\left ( m^{2}-1 \right ) = \left(\frac{1}{\cos\theta}+\frac{1}{\sin\theta}\right)\left((\cos\theta+\sin\theta)^{2}-1\right)$
$⇒n\left ( m^{2}-1 \right ) = \left(\frac{1}{\cos\theta}+\frac{1}{\sin\theta}\right)\left(\cos^{2}\theta+2\sin\theta\cos\theta+\sin^{2}\theta-1\right)$
We know that $\cos^{2}\theta+\sin^{2}\theta=1$,
$⇒n\left ( m^{2}-1 \right ) = \left(\frac{1}{\cos\theta}+\frac{1}{\sin\theta}\right)\left(2\sin\theta\cos\theta\right)$
$⇒n\left ( m^{2}-1 \right ) = 2\sin\theta\cos\theta\left(\frac{1}{\cos\theta}+\frac{1}{\sin\theta}\right)$
$⇒n\left ( m^{2}-1 \right ) = 2\left(\sin\theta+\cos\theta\right)$
But we know that $\sin\theta+\cos\theta=m$,
$⇒n\left ( m^{2}-1 \right ) = 2m$
Hence, the correct answer is $2m$.