Question

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

• A. True
• B. False

Answer 1

The correct option is A True

$\Rightarrow$$(\cot \theta +\tan \theta )=m$

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

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

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

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

Similarly,

$\Rightarrow$$n=\frac{{\sin }^2\theta }{\cos \theta }$

Therefore,

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

$={(\frac{1}{{\sin }^2\theta {\cos }^2\theta }\cdot \frac{{\sin }^2\theta }{\cos \theta })}^{2/3}-{(\frac{{\sin }^4\theta }{{\cos }^2\theta }\cdot \frac{1}{\sin \theta \cos \theta })}^{2/3}$

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

$=1$

$=$ R.H.S.

Hence proved.