### If $sin {\space} \theta + cos {\space} \theta = m$ and $sec {\space} \theta + cosec {\space} \theta = n$, prove that $n(m^2 - 1) = 2m.$

1. Substituting $sin {\space} \theta + cos {\space} \theta = m$ and $sec {\space} \theta + cosec {\space} \theta = n$ in $n(m^2 - 1),$ we have \begin{aligned} n(m^2 -1) &= sec {\space} \theta + cosec {\space} \theta \space ((sin {\space} \theta + cos {\space} \theta)^2 - 1) \\ &= (sec {\space} \theta + cosec {\space} \theta)(sin^2 {\space} \theta + cos^2 {\space} \theta + 2sin{\space} \theta cos {\space} \theta - 1) \\ &= \bigg( \dfrac { 1 } { cos {\space} \theta } + \dfrac { 1 } { sin {\space} \theta } \bigg) ( 1 + 2sin{\space} \theta cos {\space} \theta - 1 ) &&[\because sin^2 {\space} \theta + cos^2 {\space} \theta = 1, \\ & && sec {\space} \theta = \dfrac { 1 } { cos {\space} \theta }, \\ & &&\text{ and } cosec {\space} \theta = \dfrac { 1 } { sin {\space} \theta }] \\ &= \bigg( \dfrac { sin {\space} \theta + cos {\space} \theta } { sin {\space} \theta cos {\space} \theta } \bigg) ( 2sin{\space} \theta cos {\space} \theta ) \\ &= 2( sin {\space} \theta + cos {\space} \theta) = 2m \end{aligned}
2. Hence, $n(m^2 -1)$ = $2m$.