Question

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

Answer 1

$\cot \theta +\tan \theta =m$

$=\frac{1+{\tan }^2\theta }{\tan \theta }=\frac{{\sec }^2\theta }{\tan \theta }=\frac{1}{{\cos }^2\theta }\times \frac{\cos \theta }{\sin \theta }$

$=\frac{1}{\sin \theta \cos \theta }=m$

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

$m^2=\frac{1}{{\sin }^2\theta {\cos }^2\theta },n^2=\frac{{\sin }^4\theta }{{\cos }^2\theta }$

$m^{2}n=\frac{1}{{\sin }^2\theta {\cos }^2\theta }\times \frac{{\sin }^2\theta }{\cos \theta }=\frac{1}{{\cos }^3\theta }={\sec }^3\theta$

$n^{2}m=\frac{{\sin }^4\theta }{{\cos }^2\theta }\times \frac{1}{\sin \theta \cos \theta }={\tan }^3\theta$

${\left(m^{2}n\right)}^{2/3}-{\left(m{n}^2\right)}^{2/3}$

$={\left({\sec }^3\theta \right)}^{2/3}-{\left({\tan }^3\theta \right)}^{2/3}$

$={\sec }^2\theta -{\tan }^2\theta =1$