Question

If $\text{cosec}\theta -\sin \theta =m$ and $\sec \theta -\cos \theta =n$, then prove that $(m^{2}n{)}^{2/3}+(m{n}^2{)}^{2/3}=1$

Answer 1

Given, $\text{cosec}\theta -\sin \theta =m,\sec \theta -\cos \theta =n$

Thus $m=\frac{1-{\sin }^2\theta }{\sin \theta }=\frac{{\cos }^2\theta }{\sin \theta }$

$n=\frac{1-{\cos }^2\theta }{\cos \theta }=\frac{{\sin }^2\theta }{\cos \theta }$

$m^{2}n={\cos }^3\theta$ => $(m^{2}n{)}^{2/3}={\cos }^2\theta$

$m{n}^2={\sin }^3\theta =>(m{n}^2{)}^{2/3}={\sin }^2\theta$

$m{n}^2+m^{2}n={\cos }^2\theta +{\sin }^2\theta =1$

Thus any power to digit $1$ will always be equal to $1$.