How prove this idenity this $mv-3nu=m-3u$ with unit circle 
Assmue the $m,n,u,v$ be real numbers,and such
  $$m^2+n^2=1,u^2+v^2=1,nv>0,m>0,u>0$$
  and 
  $$5mu=3(1-nv)$$
show that
  $$mv-3nu=m-3u$$

Following is My methods:
let
$$m=\cos{x},n=\sin{x},u=\cos{y},v=\sin{y}$$
and
$$5mu=3(1-nv)\Longrightarrow 5\cos{x}\cos{y}=3(1-\sin{x}\sin{y})$$
$$\Longrightarrow 5\cos{x}\cos{y}+3\sin{x}\sin{y}=3$$
$$4\cos{(x-y)}+\cos{(x+y)}=3\Longrightarrow -8\sin^2{\dfrac{x-y}{2}}+4+2\cos^2{\dfrac{x+y}{2}}-1=3$$
$$\Longrightarrow \cos{\dfrac{x+y}{2}}=2\sin{\dfrac{x-y}{2}}$$
$$\Longrightarrow 1-\tan{\dfrac{x}{2}}\tan{\dfrac{y}{2}}=2\tan{\dfrac{x}{2}}-2\tan{\dfrac{y}{2}}$$
then we must show that
$$\Longleftrightarrow \cos{x}\sin{y}-3\sin{x}\cos{y}=\cos{x}-3\cos{y}$$
$$\Longleftrightarrow \cos{x}(1-\sin{y})=3\cos{y}(1-\sin{x})$$
$$\Longleftrightarrow\dfrac{\cos{x}}{1-\sin{x}}=3\dfrac{\cos{y}}{1-\sin{y}}\Longleftrightarrow \dfrac{1+\tan{\frac{x}{2}}}{1-\tan{\frac{x}{2}}}=3\dfrac{1+\tan{\frac{y}{2}}}{1-\tan{\frac{y}{2}}}\tag{2}$$
$$\Longleftrightarrow 1-\tan{\dfrac{x}{2}}\tan{\dfrac{y}{2}}=2\tan{\dfrac{x}{2}}-2\tan{\dfrac{y}{2}}$$
so it is clear
My Question:have without this trigonometric methods?such this famous Lagrange's identity
$$(a^2+b^2)(c^2+d^2)=(ac+bd)^2+(ad-bc)^2$$
since $m^2+n^2=1,u^2+v^2=1$
$$(5mu+3nv)=3\Longrightarrow (mv+3u)=3nu+m$$

It follows also directly by substituting the equations. Let $u= - 3(nv - 1)/(5m)$. Then the desired equation is given by
$$
9(nv - 1)(n - 1) + 5(v - 1)m^2=0.
$$
Now substitute $m^2:=1-n^2$. Then we need to show
$$
(4nv + 5n - 5v - 4)(n-1)=0.
$$
Since $n-1\neq 0$ because of $m>0$ we need to show $4nv + 5n - 5v - 4=0$.
This follows from $u^2+v^2=1$, which is equivalent to
$$
(4nv + 5n - 5v - 4)(4nv-5n+5v-4)=0.
$$
Here the second factor cannot be zero. Assume it is zero, i.e., $v=(5n+4)/(4n+5)$.
Then $u=-3(n-1)(n+1)/(4n+5)m)<0$, a contradiction. It follows that the first factor is zero, i.e., 
$$
4nv + 5n - 5v - 4=0.
$$
