Question

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

Answer 1

Since $\text{cosec}\theta -\sin \theta =m$
and $\sec \theta -\cos \theta =n$
$\Rightarrow \frac{1-si{n}^2\theta }{\sin \theta }=m,\frac{1-{\cos }^2\theta }{\cos \theta }=n$
$\Rightarrow \frac{{\cos }^2\theta }{\sin \theta }=m\Rightarrow \frac{{\sin }^2\theta }{\cos \theta }=n$
Now, $m^{2}n={\cos }^3\theta$
$\Rightarrow \cos \theta =(m^{2}n{)}^{\frac{1}{3}}....(1)$
Also, $m{n}^2={\sin }^3\theta$
$\Rightarrow \sin \theta =(m{n}^2{)}^{\frac{1}{3}}....(2)$
Squaring and adding $\left(1\right)$ and $\left(2\right)$, we have
$(m^{2}n{)}^{\frac{2}{3}}+(m{n}^2{)}^{\frac{2}{3}}=1$