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

\(\displaystyle \sin \theta+\cos \theta=p, \sec \theta+\operatorname{cosec} \theta=q\) \[\begin{aligned} \text { LHS }= & \mathrm{q}\left(\mathrm{p}^{2}-1\right) \\ = & (\sec \theta+\operatorname{cosec} \theta)\left[(\boldsymbol{\operatorname { s i n }} \theta+\cos \theta)^{2}-1\right] \\ = & {\left[\frac{\mathbf{1}}{\boldsymbol{\operatorname { c o s }} \theta}+\frac{\mathbf{1}}{\boldsymbol{\operatorname { s i n }} \theta}\right]\left[\boldsymbol{\operatorname { s i n }}^{2} \theta+\cos ^{2} \theta+2 \sin \theta \cos \theta-1\right] } \\ = & \left(\frac{\sin \theta+\cos \theta}{\cos \theta \sin \theta}\right)[1+2 \sin \theta \cos \theta-1] \\ & =\frac{(\sin \theta+\cos \theta)}{\cos \theta \sin \theta}(2 \sin \theta \cos \theta) \\ & =2(\sin \theta+\cos \theta) \\ & =2 \mathrm{p}=\mathrm{RHS} \end{aligned} \]