Question

If $\cot \theta +\tan \theta =x$, $\sec \theta -\cos \theta =y$, eliminate $\theta$.

Answer 1

$\left(\frac{1}{\tan \theta }\right)+\tan \theta =x$, $\left(\frac{1}{\cos \theta }\right)+\cos \theta =y$
or $\frac{{\sec }^2\theta }{\tan \theta }=x$, $\frac{{\sin }^2\theta }{\cos \theta }=y$.
$x=\frac{1}{\sin \theta \cos \theta }$, $y=\frac{{\sin }^2\theta }{\cos \theta }$
$\therefore x^{2}y=\frac{1}{{\cos }^3\theta }={\sec }^3\theta$, $x^{2}y={\tan }^3\theta$
Since ${\sec }^2\theta -{\tan }^2\theta =1$
$\therefore {\left(x^{2}y\right)}^{\frac{2}{3}}-{\left(x{y}^2\right)}^{\frac{2}{3}}=1$