Eliminate $\theta$ 
Eliminate $\theta$ in
$$\sin \theta + \mbox{cosec} \, \theta = m$$
$$\sec \theta - \cos \theta = n$$

My approach-
I multiplied the first equation by $\sin \theta$ and the second equation by $\cos \theta$ but it doesn't give me the desired answer..
 A: Squaring and adding,we get
$1+cosec^2\theta+\sec^2\theta=m^2+n^2.....(1)$
We need to find $cosec\theta$ and $\sec\theta$ from the given equations
First equation becomes $\frac{1}{\csc\theta}+\csc\theta=m$
$cosec^2\theta-m$ $cosec\theta+1=0$
Similarly second equation becomes
$\sec^2\theta-n\sec\theta-1=0$
Solving these using quadratic formula,
$cosec\theta=\frac{m\pm\sqrt{m^2-4}}{2}$
$\sec\theta=\frac{n\pm\sqrt{n^2+4}}{2}$
Put these values in equation $(1)$.
A: We have
$$\frac{1}{\cos (\theta)} - \cos (\theta) = n \qquad \qquad \qquad \qquad \sin (\theta) + \frac{1}{\sin(\theta)} = m$$
Let $x := \cos (\theta)$ and $y := \sin(\theta)$. Hence, $x^2 + y^2 = 1$. The two equations above can be rewritten as
$$\frac{1}{x} - x = n \qquad \qquad \qquad \qquad y + \frac{1}{y} = m$$
or, as follows
$$x^2 = 1 - n x \qquad \qquad \qquad \qquad y^2 = m y - 1$$
Since $x^2 + y^2 = 1$, we obtain the equation $m y - nx = 1$. Thus, we have the intersection of the unit circle and a line
$$x^2 + y^2 = 1\qquad \qquad \qquad \qquad m y - nx = 1$$
If $m, n$ are such that the intersection is not empty, then from $x, y$ we obtain the angle $\theta$. Note that the cardinality of the intersection is $0$, $1$, or $2$.
A: We have $\dfrac{1-\cos^2\theta}{\cos\theta}=n\implies\sin^2\theta=n\cos\theta\  \  \  \ (1)$
$\iff\cos^2\theta+n\cos\theta-1=0\  \ \  \  (2)$
and multiply $\sin\theta$ with the both sides of $\sin \theta +\csc\theta = m$, $$\sin^2\theta=m\sin\theta-1\  \ \  \  (3)$$
$(1),(3)\implies n\cos\theta+1=m\sin\theta$
Squaring & on rearrangement, $(m^2+n^2)\cos^2\theta+2n\cos\theta+1-m^2=0\  \ \ \ (4)$
Solve $(2),(4)$ for $\cos^2\theta,\cos\theta$ and utilize the fact $\cos^2\theta=(\cos\theta)^2$
A: using shorthand 
$$ s = \sin \theta; c = \cos \theta; s2 = \sin 2 \theta; c2 = \cos 2 \theta; $$
Square and add
$$ s^2 +1/s^2 + c^2 + 1/c^2 = m^2+n^2$$
or
$$ sc= \frac{1}{\sqrt{m^2n^2-1}}, \, s2= \frac{2}{\sqrt{m^2+n^2-1}}  $$
$$c2 = \sqrt{1-s2^2}=\sqrt{ \frac{ m^2+n^2-5}{m^2+n^2-1}}$$
$$s = \sqrt { \frac{ 1-c2}{2}}$$
Plug into first given relation
$$\sqrt { \frac{ 1-c2}{2}}  +  \sqrt { \frac{2}{ 1-c2}} = m $$
Simplify. There are two signs for every $\pm$, so there are eight eliminants to be considered.
A: Take the second equation and convert sec and cosec into sine and cos,
take 1st equation and square it on both sides to get 1+2sinxcosx = m^2
Substute m =sinx + cosx and sinxcosx = (m^2-1)*0.5 in the second and you will get the answer easily
