Question

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

Answer 1

The question should be:

prove that $(x^{2}y{)}^{\frac{2}{3}}-(x{y}^2{)}^{\frac{2}{3}}=1$

$\cot \theta +\tan \theta =\frac{1+{\tan }^2\theta }{\tan \theta }=\frac{{\sec }^2\theta }{\tan \theta }$

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

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

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

$\therefore x^2=\frac{1}{{\sin }^2\theta {\cos }^2\theta },y^2=\frac{{\sin }^4\theta }{{\cos }^2\theta }$

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

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

$\therefore (x^{2}y{)}^{\frac{2}{3}}-(x{y}^2{)}^{\frac{2}{3}}={\left({\sec }^3\theta \right)}^{\frac{2}{3}}-{\left({\tan }^3\theta \right)}^{\frac{2}{3}}$

$={\sec }^2\theta -{\tan }^2\theta$

$=1$