Question

If (cot θ + tan θ) = m and (sec θ − cos θ) = n, prove that (m²n)^2/3 − (mn²)^2/3 = 1.

Answer 1

$\begin{array}{l}\\ \text{We have}(\cot \theta +\tan \theta )=m\text{and}(\sec \theta -\text{cos}\theta )=n\\ \\ \text{Now,}{m}^{2}n=\left[{(\cot \theta +\tan \theta )}^2(\sec \theta -\text{cos}\theta )\right]\\ \\ \text{=}\left[{\left(\frac{1}{\tan \theta }+\tan \theta \right)}^2\left(\frac{1}{\cos \theta }-\cos \theta \right)\right]\\ =\frac{{(1+{\tan }^2\theta )}^2}{{\tan }^2\theta }\times \frac{(1-{\cos }^2\theta )}{\cos \theta }\\ =\frac{{\sec }^4\theta }{{\tan }^2\theta }\times \frac{{\sin }^2\theta }{\cos \theta }\\ =\frac{{\sec }^4\theta }{\frac{{\sin }^2\theta }{{\cos }^2\theta }}\times \frac{{\sin }^2\theta }{\cos \theta }\\ =\frac{{\cos }^2\theta \times {\sec }^4\theta }{\cos \theta }\\ =\cos \theta {\sec }^4\theta \\ =\frac{1}{\sec \theta }\times {\sec }^4\theta ={\sec }^3\theta \\ \therefore {\left(m^{2}n\right)}^{\frac{2}{3}}={\left({\sec }^3\theta \right)}^{\frac{2}{3}}={\sec }^2\theta \\ \end{array}$

$\begin{array}{l}\text{Again,}m{n}^2=\left[(\cot \theta +\tan \theta ){(\sec \theta -\text{cos}\theta )}^2\right]\\ \\ \text{=}[(\frac{1}{\tan \theta }+\tan \theta ).{(\frac{1}{\cos \theta }-\cos \theta )}^2]\\ =\frac{(1+{\tan }^2\theta )}{\tan \theta }\times \frac{{(1-{\cos }^2\theta )}^2}{{\cos }^2\theta }\\ =\frac{{\sec }^2\theta }{\tan \theta }\times \frac{{\sin }^4\theta }{{\cos }^2\theta }\\ =\frac{{\sec }^2\theta }{\frac{\sin \theta }{\cos \theta }}\times \frac{{\sin }^4\theta }{{\cos }^2\theta }\\ =\frac{{\sec }^2\theta \times {\sin }^3\theta }{\cos \theta }\\ =\frac{1}{{\cos }^2\theta }\times \frac{{\sec }^3\theta }{\cos \theta }={\tan }^3\theta \\ \therefore {\left(m{n}^2\right)}^{\frac{2}{3}}={\left({\tan }^3\theta \right)}^{\frac{2}{3}}={\tan }^2\theta \\ \text{Now,}{\left(m^{2}n\right)}^{\frac{2}{3}}-{\left(m{n}^2\right)}^{\frac{2}{3}}\\ ={\sec }^2\theta -{\tan }^2\theta =1\\ =\text{RHS}\\ \text{Hence proved}\text{.}\end{array}$