Prove that $m^{2} + n^{2} = \csc^{2}\left(\theta\right)$ $$
\mbox{If}\quad\left\lbrace%
\begin{array}{rcl} 
m^{2} + m'^{2} + 2mm'\cos\left(\theta\right)  & = & 1
\\
n^{2} + n'^{2} + 2nn'\cos\left(\theta\right)  & = & 1
\\
mn + m'n' + (mn' + m'n)\cos\left(\theta\right) & = & 0
\end{array}\right\rbrace
\quad\mbox{then prove that}\ m^{2} + n^{2}  = \csc^{2}\left(\theta\right)
$$
Don't know how go about this. Please help.
 A: Here's a geometric proof:
Let $\alpha=\pi-\theta$. Then
$$
1 = m^2 + m'^2 - 2mm'\cos\alpha \tag{1}
$$
is the cosine rule in a triangle with sides $m$, $m'$ and $1$, and
$$
1 = n^2 + n'^2 - 2nn'\cos\alpha \tag{2}
$$
is the cosine rule in a triangle with sides $n$, $n'$ and $1$. The third equation is
$$
0 = 2mn + 2m'n' - 2(mn' + m'n)\cos\alpha.
$$
When we add these three equations side by side, we get
$$
2 = m^2 + n^2 + 2mn + m'^2 + n'^2 + 2m'n' - 2(mm' + nn' + mn' + m'n)\cos\alpha
$$
or
$$
2= (m+n)^2 + (m'+n')^2 - 2(m+n)(m'+n')\cos\alpha, \tag{3}
$$
which is the cosine rule in a triangle with sides $m+n$, $m'+n'$ and $\sqrt{2}$. We can combine all three triangles in a single figure:

Now, we see that the three pink triangles are equal: they are right triangles with hypotenuse $1$ and sides 
$$x=m\sin\alpha=m\sin\theta$$ 
and 
$$y=n\sin\alpha=n\sin\theta,$$
so that
$$
m^2\sin^2\theta + n^2\sin^2\theta = 1,
$$
or
$$
m^2 + n^2 = \csc^2\theta.
$$
A: For convenience, let $a=m$, $b=m^{\prime}$, $x=n$, $y=n^{\prime}$; so
$a^2+b^2+2ab\cos\theta=1$, $\;\;x^2+y^2+2xy\cos\theta=1$, $\;\;ax+by+(ay+bx)\cos\theta=0$.
From the 3rd equation, $(a+b\cos\theta)x+(b+a\cos\theta)y=0$, so $
y=-\frac{a+b\cos\theta}{b+a\cos\theta}x$.
Substituting into the 2nd equation gives $x^2+\left(\frac{a+b\cos\theta}{b+a\cos\theta}\right)^2x^2-\frac{2(a+b\cos\theta)\cos\theta}{b+a\cos\theta}x^2=1$, so
$\frac{(b+a\cos\theta)^2+(a+b\cos\theta)^2-2(b+a\cos\theta)(a+b\cos\theta)\cos\theta}{(b+a\cos\theta)^2} x^2=1$.
Then $x^2=\frac{(b+a\cos\theta)^2}{b^2+2ab\cos\theta+a^2\cos^{2}\theta+a^2+2ab\cos\theta+b^2\cos^{2}\theta-2ab\cos\theta-2a^2\cos^{2}\theta-2b^2\cos^{2}\theta-2ab\cos^{3}\theta}$, so 
$x^2=\frac{(b+a\cos\theta)^2}{a^2+b^2-a^2\cos^{2}\theta-b^2\cos^{2}\theta+2ab\cos\theta(1-\cos^{2}\theta)}=\frac{(b+a\cos\theta)^2}{(a^2+b^2+2ab\cos\theta)\sin^{2}\theta}=\frac{(b+a\cos\theta)^2}{\sin^2\theta}$ using Equation 1.
Then $a^2+x^2=a^2+\frac{(b+a\cos\theta)^2}{\sin^2\theta}=\frac{a^2\sin^2\theta+b^2+2ab\cos\theta+a^2\cos^2\theta}{\sin^2\theta}=\frac{a^2+b^2+2ab\cos\theta}{\sin^2\theta}=\frac{1}{\sin^2\theta}=\csc^2\theta$,
(again using Equation 1).
A: Using the same set of notations used by user84413,
$\displaystyle ax+by+(ay+bx)\cos\theta=0\  \ \  \  (1)$
$\displaystyle a^2+b^2+2ab\cos\theta=1\iff (b+a\cos\theta)^2=1-a^2\sin^2\theta\  \ \ \ \ (2)$
Similarly, $\displaystyle x^2+y^2+2xy\cos\theta=1\iff(y+x\cos\theta)^2=1-x^2\sin^2\theta\  \ \ \ \ (3)$
Multiplying $(2),(3)$
$\displaystyle(1-a^2\sin^2\theta)(1-x^2\sin^2\theta)=\{(b+a\cos\theta)(y+x\cos\theta)\}^2$
$\displaystyle\implies 1-(a^2+x^2)\sin^2\theta+a^2x^2\sin^4\theta=\{by+\cos\theta(ay+bx)+ax\cos^2\theta\}^2$
Now putting the value of $by+(ay+bx)\cos\theta$ from $(1),$
$\displaystyle\{by+\cos\theta(ay+bx)+ax\cos^2\theta\}^2=\{(-ax)+ax\cos^2\theta\}^2=a^2x^2(-\sin^2\theta)^2=a^2x^2\sin^4\theta$
$\displaystyle\implies 1-(a^2+x^2)\sin^2\theta+a^2x^2\sin^4\theta=a^2x^2\sin^4\theta$
Now simplify. 
