If (cot θ + tan θ) = m and (sec θ − cos θ) = n,

Solution:

We have $(\cot \theta+\tan \theta)=m$ and $(\sec \theta-\cos \theta)=n$

Now, $m^{2} n=\left[(\cot \theta+\tan \theta)^{2}(\sec \theta-\cos \theta)\right]$

$=\left[\left(\frac{1}{\tan \theta}+\tan \theta\right)^{2}\left(\frac{1}{\cos \theta}-\cos \theta\right)\right]$

$=\frac{\left(1+\tan ^{2} \theta\right)^{2}}{\tan ^{2} \theta} \times \frac{\left(1-\cos ^{2} \theta\right)}{\cos \theta}$

$=\frac{\sec ^{4} \theta}{\tan ^{2} \theta} \times \frac{\sin ^{2} \theta}{\cos \theta}$

$=\frac{\sec ^{4} \theta}{\frac{\sin ^{2} \theta}{\cos ^{2} \theta}} \times \frac{\sin ^{2} \theta}{\cos \theta}$

$=\frac{\cos ^{2} \theta \times \sec ^{4} \theta}{\cos \theta}$

$=\cos \theta \sec ^{4} \theta$

$=\frac{1}{\sec \theta} \times \sec ^{4} \theta=\sec ^{3} \theta$

$\therefore\left(m^{2} n\right)^{\frac{2}{3}}=\left(\sec ^{3} \theta\right)^{\frac{2}{3}}=\sec ^{2} \theta$

Again, $m n^{2}=\left[(\cot \theta+\tan \theta)(\sec \theta-\cos \theta)^{2}\right]$

$=\left[\left(\frac{1}{\tan \theta}+\tan \theta\right) \cdot\left(\frac{1}{\cos \theta}-\cos \theta\right)^{2}\right]$

$=\frac{\left(1+\tan ^{2} \theta\right)}{\tan \theta} \times \frac{\left(1-\cos ^{2} \theta\right)^{2}}{\cos ^{2} \theta}$

$=\frac{\sec ^{2} \theta}{\tan \theta} \times \frac{\sin ^{4} \theta}{\cos ^{2} \theta}$

$=\frac{\sec ^{2} \theta}{\frac{\sin \theta}{\cos \theta}} \times \frac{\sin ^{4} \theta}{\cos ^{2} \theta}$

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

$=\frac{1}{\cos ^{2} \theta} \times \frac{\sec ^{3} \theta}{\cos \theta}=\tan ^{3} \theta$

$\therefore\left(m n^{2}\right)^{\frac{2}{3}}=\left(\tan ^{3} \theta\right)^{\frac{2}{3}}=\tan ^{2} \theta$

Now, $\left(m^{2} n\right)^{\frac{2}{3}}-\left(m n^{2}\right)^{\frac{2}{3}}$

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

$=\mathrm{RHS}$

Hence proved.