Question

If $\sin \theta+\cos \theta=m$ and $\sec \theta+\operatorname{cosec} \theta=n$, then $n(m+1)(m-1)$ is equal to

(A) $m$

(B) $n$

(C) $2 m$

(D) $2 n$

Answer 1

Given,$\sin \theta+\cos \theta=m$

$\begin{array}{l}\sec \theta+\operatorname{cosec} \theta=n \\n(m+1)(m-1)=n\left(m^2-1\right) \\=(\sec \theta+\operatorname{cosec} \theta) 2 \sin \theta \cos \theta \\\left(\because m^2=1+2 \sin \theta \cos \theta\right) \\=\frac{\sin \theta+\cos \theta}{\sin \theta \cos \theta} \cdot 2 \sin \theta \cos \theta=2 m \\\end{array}$

So, The correct option will be (C).