If $\displaystyle \sin \theta+\cos \theta=\mathrm{p}$ and $\displaystyle \sec \theta+\operatorname{cosec} \theta=\mathrm{q}$, then prove that $\displaystyle \mathrm{q}\left(\mathrm{p}^{2}-1\right)=2 \mathrm{p}$.

L.H.S \(\displaystyle =\mathrm{q}\left(p^{2}-1\right)\) \[\begin{aligned} & =(\sec \theta+\operatorname{cosec} \theta)\left[(\sin \theta+\cos \theta)^{2}-1\right] \\ & =\left[\frac{1}{\cos \theta}+\frac{1}{\sin \theta}\right]\left[\sin ^{2} \theta+\cos ^{2} \theta+2 \sin \theta \cos \theta-1\right] \\ & =\left[\frac{\sin \theta+\cos \theta}{\sin \theta \cos \theta}\right][1+2 \sin \theta \cos \theta-1] \\ & =2(\sin \theta+\cos \theta) \\ & =2 p \\ & =\text { R.H.S } \end{aligned} \]