If $$\operatorname{cosec}\theta-\operatorname{sin}\theta=m$$ and $$\operatorname{sec}\theta-\operatorname{cos}\theta=n$$ then eliminate $\theta$.
My work:
Squaring both the equations and then adding them will yeild $$\operatorname{cosec}^2\theta+\operatorname{sec}^2\theta=m^2+n^2+3$$ or
$$\operatorname{cos}^2\theta\operatorname{sin}^2\theta=\frac{1}{m^2+n^2+3}$$ or$$\operatorname{sin}^2 2\theta=\frac{4}{m^2+n^2+3}$$ or
$$\theta=\frac12\cdot\operatorname{arcsin}\left(\frac{2}{\sqrt{m^2+n^2+3}}\right)$$
Now to eliminate $\theta$, we will have to put its value in any of the two equations.
But actually I don't know how to evaluate inverse trigonometric values, like that of $\theta$ in terms of $m$ and $n$. And after that I guess it will be difficult to simplify the expression.
Any help is greatly appreciated.
Best Answer
Writing $x$ for $\theta,$
$$m=\csc x-\sin x=\dfrac{\cos^2x}{\sin x}$$
$$\text{Similarly, } n=\sec x-\cos x=\dfrac{\sin^2x}{\cos x}$$
$$\dfrac mn=\dfrac{\cos^3x}{\sin^3x} \implies\dfrac m{\cos^3x}=\dfrac n{\sin^3x}=\dfrac{m^{2/3}+n^{2/3}}1$$
Replacing the values of $\cos x,\sin x;$
$$ m=\dfrac{\left(\dfrac m{m^{2/3}+n^{2/3}}\right)^{2/3}}{\left(\dfrac n{m^{2/3}+n^{2/3}}\right)^{1/3}} \implies m(m^{2/3}+n^{2/3})^{1/3}n^{1/3}=m^{2/3}$$
Now I leave the rest for the OP