Question

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

Answer 1

$m^{2}n=(cosec\theta -\sin \theta {)}^2(\sec \theta -\cos \theta )={\left(\frac{1-{\sin }^2\theta }{\sin \theta }\right)}^2\left(\frac{1-{\cos }^2\theta }{\cos \theta }\right)=\frac{{\cos }^4\theta {\sin }^2\theta }{{\sin }^2\theta \cos \theta }$

$\Rightarrow \left(m^{2}n\right)={\cos }^3\theta$

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

Similarly $m{n}^2=(cosec\theta -\sin \theta )(\sec \theta -\cos \theta {)}^2=\frac{{\sin }^4\theta {\cos }^2\theta }{{\cos }^2\theta \sin \theta }={\sin }^3\theta$

$\Rightarrow (m^{2}n{)}^{2/3}+(m{n}^2{)}^{2/3}={\sin }^2\theta +{\cos }^2\theta =1$.