Question

If $\cos \theta -\sin \theta =m$, $\sec \theta -\cos \theta =n$ eliminate $\theta$.
If $\cos \theta -\sin \theta =a^3$, then
$a^2{b}^2(a^2+b^2)=1$.

Answer 1

$\cos \theta -\sin \theta =m$ and $\sec \theta -\cos \theta =n$.

or $\left(1/\sin \theta \right)-\sin \theta =m$ and $\left(1/\cos \theta \right)-\cos \theta =n$

or $\frac{{\cos }^2\theta }{\sin \theta }=m$ and $\frac{{\sin }^2\theta }{\cos \theta }=n$.

$mn=\cos \theta \sin \theta$ and ${\cos }^2\theta =m\sin \theta$.

$\therefore {\cos }^3\theta =m\sin \theta \cos \theta =m\left(mn\right)$

$\therefore m^{2}n={\cos }^3\theta$ and

$\therefore {mn}^2={\sin }^3\theta$

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

$\therefore {\left(m^{2}n\right)}^{2/3}+{\left({mn}^2\right)}^{2/3}=1$

Another form:

Put $m=a^3$, $n=b^3$ in part $\left(a\right)$.