Question

If $\cot \theta +\tan \theta =m$ and $\sec \theta -\cos \theta =n$ then which of the following is correct

• A. $(m^{2}n{)}^{\frac{2}{3}}-(n^{2}m{)}^{\frac{2}{3}}=1$
  
• B. $m{\left(m^{2}n\right)}^{\frac{1}{3}}-n{\left(m{n}^2\right)}^{\frac{1}{3}}=1$
• C. $m{\left(m{n}^2\right)}^{\frac{1}{3}}-m{\left(m{n}^2\right)}^{\frac{1}{3}}=1$
• D. $m{\left(m^{2}n\right)}^{\frac{1}{3}}-m{\left(m^{2}n\right)}^{\frac{1}{3}}=1$

Answer 1

The correct option is A

$(m^{2}n{)}^{\frac{2}{3}}-(n^{2}m{)}^{\frac{2}{3}}=1$

Given,

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

$\Rightarrow \frac{1+{\tan }^2\theta }{\tan \theta }=m$

$\Rightarrow \frac{{\sec }^2\theta }{\tan \theta }=m$

$\Rightarrow \frac{1}{\sin \theta \cos \theta }=m$........(1)

$\Rightarrow \frac{1}{{\sin }^2\theta {\cos }^2\theta }=m^2$........(2)

$\sec \theta -\cos \theta =n$

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

$\frac{{\sin }^2\theta }{\cos \theta }=n$...........(3)

$\frac{{\sin }^4\theta }{{\cos }^2\theta }=n^2$...........(4)

using (1), (2), (3) and (4), we get,

$m^{2}n-n^{2}m={\sec }^3\theta -{\tan }^3\theta$

$\Rightarrow (m^{2}n{)}^{\frac{2}{3}}-(n^{2}m{)}^{\frac{2}{3}}=({\sec }^3\theta {)}^{\frac{2}{3}}-({\tan }^3\theta {)}^{\frac{2}{3}}$

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