Question

If $cosec\theta -\sin \theta =m$, $\sec \theta -\cos \theta =n$ then $(m^{2}n{)}^{\frac{2}{3}}+(m{n}^2{)}^{\frac{2}{3}}=$

• A. 0
• B. 1
• C. -1
• D. 2

Answer 1

Answer: (B)

1

Given,

$\csc \theta -\sin \theta =m$

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

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

$\Rightarrow \frac{{\cos }^4\theta }{{\sin }^2\theta }=m^2$

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

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

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

$\Rightarrow \frac{{\sin }^4\theta }{{\cos }^2\theta }=n^2$

$m^{2}n={\cos }^3\theta$

$m{n}^2={\sin }^3\theta$

Given,

$(m^{2}n{)}^{\frac{2}{3}}+(m{n}^2{)}^{\frac{2}{3}}$

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

$={\cos }^2\theta +{\sin }^2\theta$

$=1$