If $cosec {\space} \theta - sin {\space} \theta = m$ and $sec {\space} \theta - cos {\space} \theta = n$, prove that $(m^2n)^\frac { 2 } { 3 } + (mn^2)^\frac { 2 } { 3 } = 1.$

Step by Step Explanation:
1. Let us first calculate the value of $m^2n$ and $mn^2$.
\begin{aligned} m^2n &= (cosec {\space} \theta - sin {\space} \theta)^2 . (sec {\space} \theta - cos {\space} \theta) \\ &= \bigg( \dfrac { 1 } { sin {\space} \theta} - sin {\space} \theta \bigg)^2 . \bigg( \dfrac { 1 } { cos {\space} \theta} - cos {\space} \theta \bigg) \\ &= \dfrac { ( 1 - sin^2{\space} \theta )^2 } { sin^2{\space} \theta } . \dfrac { ( 1 - cos^2{\space} \theta ) } { cos{\space} \theta }\\ &= \dfrac { (cos^2{\space} \theta )^2 } { sin^2{\space} \theta } . \dfrac { sin^2{\space} \theta } { cos{\space} \theta } \space\space\space\space\space\space\space\space[\because 1 - sin^2{\space} \theta = cos^2{\space} \theta {\space} and {\space} 1 - cos^2{\space} \theta = sin^2{\space} \theta ] \\ &= \dfrac { cos^4{\space} \theta } { sin^2{\space} \theta } {\times} \dfrac { sin^2{\space} \theta } { cos{\space} \theta }\\ &= cos^3{\space} \theta \\ \therefore {\space}{\space}{\space} &(m^2n)^\frac { 1 } { 3 } = cos{\space} \theta \space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\ldots \text{(i)} \end{aligned}
2. Now,
\begin{align} mn^2 &= (cosec {\space} \theta - sin {\space} \theta) . (sec {\space} \theta - cos {\space} \theta)^2 \\ &= \bigg( \dfrac { 1 } { sin {\space} \theta} - sin {\space} \theta \bigg) . \bigg( \dfrac { 1 } { cos {\space} \theta} - cos {\space} \theta \bigg)^2 \\ &= \dfrac { ( 1 - sin^2{\space} \theta ) } { sin {\space} \theta } . \dfrac { ( 1 - cos^2{\space} \theta )^2 } { cos^2{\space} \theta }\\ &= \dfrac { cos^2{\space} \theta } { sin {\space} \theta }{\times}\dfrac { (sin^2{\space} \theta)^2 } { cos^2{\space} \theta } \space\space\space\space\space\space\space\space [\because 1 - sin {\space} \theta = cos^2{\space} \theta {\space} and {\space} 1 - cos^2{\space} \theta = sin^2{\space} \theta ] \\ &= sin^3{\space} \theta \\ \therefore {\space}{\space}{\space} &(mn^2)^\frac { 1 } { 3 } = sin{\space} \theta \space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\ldots \text{(ii)} \end{align}
3. On squaring and adding $\text{(i)}$ and $\text{(ii)}$, we get \begin{aligned} (m^2n)^\frac { 2 } { 3 } + (mn^2)^\frac { 2 } { 3 } = (cos{\space} \theta)^2 + (sin{\space} \theta)^2 = 1 \space\space\space\space\space\space\space\space\space\space[\because cos^2{\space} \theta + sin^2{\space} \theta = 1] \end{aligned}
4. Hence, $(m^2n)^\frac { 2 } { 3 } + (mn^2)^\frac { 2 } { 3 } = \bf 1.$