Question

If $\cot \theta +\tan \theta =x$ and $\sec \theta -\cos \theta =y$, then prove that
${\left(x^{2}y\right)}^{2/3}-{\left(x{y}^2\right)}^{2/3}=1$

Answer 1

$\frac{\cos \theta }{\sin \theta }+\frac{\sin \theta }{\cos \theta }=x$

$\Rightarrow \frac{1}{\sin \theta \cos \theta }=x$

$y=\frac{1}{\cos \theta }-\cos \theta =\frac{1-{\cos }^2\theta }{\cos \theta }=\frac{{\sin }^2\theta }{\cos \theta }$

$x^{2}y=\frac{{\sin }^2\theta }{\cos \theta }\times \frac{1}{{\sin }^2\theta {\cos }^2\theta }=\frac{1}{{\cos }^3\theta }$

$x{y}^2=\frac{1}{\sin \theta \cos \theta }\times \frac{{\sin }^4\theta }{{\cos }^2\theta }=\frac{{\sin }^3\theta }{{\cos }^3\theta }$

$\Rightarrow (x^{2}y{)}^{2/3}-(x{y}^2{)}^{2/3}=\frac{1}{{\cos }^2\theta }-\frac{{\sin }^2\theta }{{\cos }^2\theta }=1$.