Question

If $cosec\theta -\sin \theta =m$ and $\sec \theta -\cos \theta =n$, eliminate $\theta$.

• A. ${\left(m^{2}n\right)}^{4/3}+{\left(m{n}^2\right)}^{4/3}=1$
• B. ${\left(m^{4}n\right)}^{2/3}+{\left(m{n}^4\right)}^{2/3}=1$
• C. ${\left(m^{3}n\right)}^{2/3}+{\left(m{n}^3\right)}^{2/3}=1$
• D. ${\left(m^{2}n\right)}^{2/3}+{\left(m{n}^2\right)}^{2/3}=1$

Answer 1

The correct option is D ${\left(m^{2}n\right)}^{2/3}+{\left(m{n}^2\right)}^{2/3}=1$

Given $cosec\theta -\sin \theta =m$ or $\frac{1}{\sin \theta }-\sin \theta =m$
or $\frac{1-{\sin }^2\theta }{\sin \theta }=m$ or $\frac{{\cos }^2\theta }{\sin \theta }=m$ (i)
Again $\sec \theta -\cos \theta =n$ or $\frac{1}{\cos \theta }-\cos \theta =n$
or $\frac{1-{\cos }^2\theta }{\cos \theta }=n$ or
$\frac{{\sin }^2\theta }{\cos \theta }=n$ (ii)
From Eq. (i), $\sin \theta =\frac{{\cos }^2\theta }{m}$ (iii)
Putting the value of $\sin \theta$ in Eq. (ii), we get
$\frac{{\cos }^4}{m^2\cos \theta }=n$ or ${\cos }^3=m^{2}n$
$\therefore$ $\cos \theta ={\left(m^{2}n\right)}^{1/3}$ or ${\cos }^2\theta ={\left(m^{2}n\right)}^{2/3}$ (iv)
From Eq. (iii), $\sin \theta =\frac{{\cos }^2\theta }{m}=\frac{{\left(m^{2}n\right)}^{2/3}}{m}=\frac{m^{4/3}{n}^{2/3}}{m}=m^{1/3}{n}^{2/3}={\left(m{n}^2\right)}^{1/3}$
$\therefore$ ${\sin }^2\theta ={\left(m{n}^2\right)}^{2/3}$ (v)
Adding Eqs. (iv) and (v), we get
${\left(m^{2}n\right)}^{2/3}+{\left(m{n}^2\right)}^{2/3}={\cos }^2\theta +{\sin }^2\theta$
or ${\left(m^{2}n\right)}^{2/3}+{\left(m{n}^2\right)}^{2/3}=1$