Eliminate ##\theta## between a pair of given equations

[brotherbobby]

Homework Statement

Given $\boxed{\quad\text{cosec}\,\theta-\sin\theta=m\quad,\quad \sec\theta-\cos\theta=n.\qquad\mathbf{\text{Eliminate}\;\boldsymbol\theta}\quad}$

Relevant Equations

1. $\sin^2\theta+\cos^2\theta=1$,

2. If $ax^2+bx+c=0,\quad x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$

Multiplying the first equation by $\sin\theta$ throughout and the second by $\cos\theta$ throughout, we get $1-\sin^2\theta=m\sin\theta\quad,\quad 1-\cos^2\theta=n\cos\theta$.

Solving the two equations, $\quad\sin\theta=\dfrac{-m+\sqrt{m^2+4}}{2}\quad,\quad \cos\theta=\dfrac{-n+\sqrt{n^2+4}}{2}.$

Squaring them, we get $\quad\sin^2\theta=\dfrac{m^2-m\sqrt{m^2+4}+2}{2}\quad,\quad \cos^2\theta=\dfrac{n^2-n\sqrt{n^2+4}+2}{2}.$

Adding them to 1 and simplyfying, $\boxed{m^2+n^2-m\sqrt{m^2+4}-n\sqrt{n^2+4}+2=0\;}$ , which is my eliminant.

Text answer : $\boxed{\quad\boldsymbol{m^{2/3}\,n^{2/3}(m^{2/3}+n^{2/3})=1}\quad}$

I cannot reduce my answer to that of the text via algebra.

Request : Where did I go wrong?

 

[fresh_42]

I don't see a mistake but I haven't verified what you did not write. I would avoid solving any quadratic equation as long as possible, and certainly won't drop one root. The terms with the negative root might lose significant information. My calculation resulted in
\begin{align*}
\operatorname{cosec} \theta - \sin \theta &=\dfrac{1}{\sin\theta}- \sin \theta=m\\
\sec\theta-\cos\theta&=\dfrac{1}{\cos\theta}-\cos\theta=n\\[8pt]
1-\sin^2\theta&=\cos^2\theta=m\sin\theta\\
1-\cos^2\theta&=\sin^2\theta=n\cos\theta\\[8pt]
m\sin\theta&=\left(\dfrac{\sin^2\theta}{n}\right)^2=\dfrac{\sin^4\theta}{n^2}\\
n\cos\theta&=\left(\dfrac{\cos^2\theta}{m}\right)^2=\dfrac{\cos^4\theta}{m^2}\\[8pt]
mn^2&=\sin^3\theta\\
nm^2&=\cos^3\theta
\end{align*}
which at least provides a third power.

 

[renormalize]

Request : Where did I go wrong?

Here's a simpler approach. Start from your two equations:$$\csc\theta-\sin\theta=m\;,\;\sec\theta-\cos\theta=n$$ and then multiply and divide those to get the two equivalent equations: $$\cos\theta\sin\theta=m\,n\;,\;\cot^{3}\theta\equiv\frac{\cos^{3}\theta}{\sin^{3}\theta}=\frac{m}{n}$$You can easily take it from there to get the textbook answer using just the identity (1) without any need for the quadratic formula (2).

 

[anuttarasammyak]

By wolfram alpha with limited resource

It seems the plots coincide for x,y>0. By algebra it is hard to understand for me that they are same.

 

[pines-demon]

By wolfram alpha with limited resource

Wolfram Alpha also says that
$$x^{2/3}y^{2/3}(x^{2/3}+y^{2/3})=1$$
has a solution for $x>0$, given by
$$ y = -\frac1{\sqrt2}\sqrt{-x^2 - x\sqrt{x^2 + 4} - \sqrt{x^2 + 4}/x - 3}$$
Hinting at some link.

 

[brotherbobby]

Thank you @fresh_42 and @renormalize, I have solved the problem using both your methods.
Yet a doubt lingers.

I eliminated $\theta$ to obtain $\boxed{m^2+n^2-m\sqrt{m^2+4}-n\sqrt{n^2+4}+2=0\;}$

The text obtained $\boxed{\quad\boldsymbol{m^{2/3}\,n^{2/3}(m^{2/3}+n^{2/3})=1}\quad}$ as the eliminant.

Clearly, it would be very hard if not impossible to reduce one to the other. Let me assume that both answers are correct.

Does that mean there can be several eliminants to a pair of given equations?

 

[renormalize]

Clearly, it would be very hard if not impossible to reduce one to the other.

Actually, it's not all that hard to reduce one to the other.
Start from the textbook answer:$$1=m^{2/3}n^{2/3}\left(m^{2/3}+n^{2/3}\right)=m^{4/3}n^{2/3}+m^{2/3}n^{4/3}\tag{1}$$and make the substitution $a=m^{2/3},b=n^{2/3}$ to get the quadratic equation:$$ab\left(a+b\right)=a^{2}b+ab^{2}=1\tag{2}$$To maintain symmetry between $a$ and $b$, solve (2) twice to find $a$ in terms of $b$ and vice versa:$$a=-\frac{b}{2}\pm\frac{\sqrt{b^{3}+4}}{2\sqrt{b}},\;b=-\frac{a}{2}\pm\frac{\sqrt{a^{3}+4}}{2\sqrt{a}}\tag{3a,b}$$(Note that there are 4 possible combinations of square-root-signs to be investigated, but for expediency I will consider only the $(+,+)$ case here.) Now multiply (3a) by $-2 b^2$ and (3b) by $-2a^2$ to arrive at:$$0=b^{3}-b^{3/2}\sqrt{b^{3}+4}+2ab^{2},\;0=a^{3}-a^{3/2}\sqrt{a^{3}+4}+2a^{2}b\tag{4a,b}$$and then eliminate $a,b$ in favor of $m,n$:$$0=n^{2}-n\sqrt{n^{2}+4}+2m^{2/3}n^{4/3},\;0=m^{2}-m\sqrt{m^{2}+4}+2m^{4/3}n^{2/3}\tag{5a,b}$$Finally, add (5a) and (5b) to get the single equation:
\begin{align}
0 & = m^{2}+n^{2}-m\sqrt{m^{2}+4}-n\sqrt{n^{2}+4}+2\left(m^{4/3}n^{2/3}+m^{2/3}n^{4/3}\right) \nonumber \\
& = m^{2}+n^{2}-m\sqrt{m^{2}+4}-n\sqrt{n^{2}+4}+2 \nonumber\tag{5}
\end{align}in view of eq.(1). QED.

 

[anuttarasammyak]

Thank you @fresh_42 and @renormalize, I have solved the problem using both your methods.
Yet a doubt lingers.

I eliminated $\theta$ to obtain $\boxed{m^2+n^2-m\sqrt{m^2+4}-n\sqrt{n^2+4}+2=0\;}$

The text obtained $\boxed{\quad\boldsymbol{m^{2/3}\,n^{2/3}(m^{2/3}+n^{2/3})=1}\quad}$ as the eliminant.

Clearly, it would be very hard if not impossible to reduce one to the other. Let me assume that both answers are correct.

Does that mean there can be several eliminants to a pair of given equations?

$\boxed{m^2+n^2-m\sqrt{m^2+4}-n\sqrt{n^2+4}+2=0\;}$ should be extended in order to cover m,n < 0
to
(a+b+c)(a+b-c)(a-b+c)(a-b-c)=0
where
a=m^2+n^2+2
b=m\sqrt{m^2+4}
c=n\sqrt{n^2+4}
This equation is reduced to
m^2n^2(m^2+n^2+3)=1
Cubing the textbook answer
m^2 n^2 [m^2+n^2+3\{m^{2/3}n^{2/3}(m^{2/3}+n^{2/3})\}]=1
{ }=1 so they are same. The plots : https://www.wolframalpha.com/input?i=plot+(x^2)^(1/3)(y^2)^(1/3)((x^2)^(1/3)+(y^2)^(1/3))=1+and++x^2y^2(x^2+y^2+3)=1&lang=en

@brotherbobby I find your way if extented to cover minus m,n cases gives naturally more ellegant answer than the textbook's.