eliminate A,B and C How to eliminate $A$, $B$, $C$ from the given three equations:
$$X \cos A+ Y \cos B + Z \cos C=0$$
$$X \sin A+ Y \sin B + Z \sin C=0$$
$$X \sec A+ Y \sec B + Z \sec C=0$$
Answer is 
$$(X^2+ Y^2 - Z^2 )^2 = 4 X^2 Y^2$$
I was squaring and adding 1st and 2nd eqn .But could not use the 3rd one.Please anybody help.
 A: Since $X\cos{A}=-Y\cos{B}-Z\cos{C}$ and $X\sin{A}=-Y\sin{B}-Z\sin{C}$, we obtain
$$X^2(\sin^2A+\cos^2A)=(Y\cos{B}+Z\cos{C})^2+(Y\sin{B}+Z\sin{C})^2$$ or
$$X^2=Y^2+Z^2+2YZ\cos(B-C).$$
In another hand, $X\cos{A}=-Y\cos{B}-Z\cos{C}$ and $\frac{X}{\cos{A}}=-\frac{Y}{\cos{B}}-\frac{Z}{\cos{C}}$, which gives
$$X^2=\left(Y\cos{B}+Z\cos{C}\right)\left(\frac{Y}{\cos{B}}+\frac{Z}{\cos{C}}\right)$$ or
$$X^2=Y^2+Z^2+YZ\left(\frac{\cos{B}}{\cos{C}}+\frac{\cos{C}}{\cos{B}}\right),$$
which gives
$$YZ\left(\frac{\cos{B}}{\cos{C}}+\frac{\cos{C}}{\cos{B}}-2\cos(B-C)\right)=0$$ or
$$YZ\left(\cos^2B+\cos^2C-2\cos^2B\cos^2C-2\cos{B}\cos{C}\sin{B}\sin{C}\right)=0$$ or
$$YZ\left(\cos^2B\sin^2{C}+\cos^2C\sin^2B-2\cos{B}\cos{C}\sin{B}\sin{C}\right)=0$$ or
$$YZ\sin^2(B-C)=0.$$
By the same way we can get $XZ\sin^2(A-C)=0$ and $XY\sin^2(A-B)=0$.
Now, if $Y=Z=0$ then $X\cos{A}=\frac{X}{\cos{A}}=0$, which gives $X=0$ and
$$(X+Y+Z)\prod_{cyc}(X+Y-Z)=0.$$
Let $Z=0$ and $Y\neq0$.
Hence, $X\cos{A}=-Y\cos{B}$ and $\frac{X}{\cos{A}}=-\frac{Y}{\cos{B}}$, which gives
$$X^2=Y^2$$ or
$$(X+Y)(X-Y)=0,$$ which with $Z=0$ gives
$$(X+Y+Z)\prod_{cyc}(X+Y-Z)=0$$ again. 
Let $XYZ\neq0$.
Hence, $$\sin(A-B)=\sin(A-C)=\sin(B-C)=0,$$ which gives
$$\sin{A}=\pm\sin{B}=\pm\sin{C}.$$
Now, if $\sin{A}=0$ then $\cos{A}=\pm1$, $\cos{B}=\pm1$ and $\cos{C}=\pm1$,
which gives $\pm X\pm Y\pm Z=0$ and from here we obtain
$$(X+Y+Z)\prod_{cyc}(X+Y-Z)=0$$ again. 
If $\sin{A}\neq0$ then from second given equation we get $X\pm Y\pm Z=0$, which gives
$$(X+Y+Z)\prod_{cyc}(X+Y-Z)=0$$ again and since $(X+Y+Z)\prod\limits_{cyc}(X+Y-Z)=0$ it's just
$$(X^2+Y^2-Z^2)^2=4X^2Y^2,$$
we are done!
