Question

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

Answer 1

$cosec\theta -\sin \theta =m\left(\text{Given}\right)$

$\Rightarrow \frac{1}{\sin \theta }-\sin \theta =m$

$\Rightarrow \frac{{\cos }^2\theta }{\sin \theta }=m$

Similarly,

$\sec \theta -\cos \theta =n\left(\text{Given}\right)$

$\Rightarrow \frac{1}{\cos \theta }-\cos \theta =n$

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

Therefore,

Taking L.H.S.-

${\left(m^{2}n\right)}^{2/3}+{\left(m{n}^2\right)}^{2/3}$

$={(\frac{{\cos }^4\theta }{{\sin }^2\theta }\cdot \frac{{\sin }^2\theta }{\cos \theta })}^{2/3}+{(\frac{{\cos }^2\theta }{\sin \theta }\cdot \frac{{\sin }^4\theta }{{\cos }^2\theta })}^{2/3}$

$={\left({\cos }^3\theta \right)}^{2/3}+{\left({\sin }^3\theta \right)}^{2/3}$

$={\cos }^2\theta +{\sin }^2\theta$

$=1$

$=$ R.H.S.

Hence proved.