Eliminate $\alpha,\beta,\gamma$ from the system of equations Eliminate $\alpha,\beta,\gamma$ from the following system of equations.
$$a\cos(\alpha)+b\cos(\beta)+c\cos(\gamma)=0$$
$$a\sin(\alpha)+b\sin(\beta)+c\sin(\gamma)=0$$
$$a\sec(\alpha)+b\sec(\beta)+c\sec(\gamma)=0$$
My try:
Squaring and adding first two equations, we get:
$$\cos(\alpha-\beta)=\frac{c^2-a^2-b^2}{2ab}$$
$$\cos(\gamma-\beta)=\frac{a^2-c^2-b^2}{2cb}$$
$$\cos(\alpha-\gamma)=\frac{b^2-a^2-c^2}{2ac}$$
Now the RHS of above all looks like negative cosines of triangle $\Delta ABC$.
But I am not sure whether it will help.
This question is taken from plane trigonometry part 1 by SL Loney book. Page number 264, question number 176.

We have the following system of equations
\begin{align}
a\cos(\alpha)+b\cos(\beta)+c\cos(\gamma) &= 0, \tag{1}\\
a\sin(\alpha)+b\sin(\beta)+c\sin(\gamma) &= 0, \tag{2}\\
a\sec(\alpha)+b\sec(\beta)+c\sec(\gamma) &= 0. \tag{3}
\end{align}
From (1) and (2), we have
$$a^2 + b^2 + 2ab \cos \alpha\cos \beta + 2ab \sin \alpha \sin \beta = c^2$$
and thus
$$(2ab)^2\sin^2 \alpha \sin^2\beta = (c^2 - a^2 - b^2 - 2ab\cos \alpha\cos \beta)^2. \tag{4}$$
From (1) and (3), we have
$$a^2 + b^2 + ab\frac{\cos \alpha}{\cos \beta} + ab \frac{\cos \beta}{\cos \alpha} = c^2$$
and thus
$$(2ab)^2(\cos^2 \alpha + \cos^2\beta) = 2\cdot (c^2 - a^2 - b^2)\cdot 2ab \cos \alpha \cos \beta. \tag{5}$$
$(4) + (5)$ gives
$$(2ab)^2(1 + \cos^2\alpha \cos^2\beta ) = (c^2 - a^2 - b^2)^2 + (2ab)^2 \cos^2 \alpha \cos^2 \beta$$
and thus
$$(2ab)^2 = (c^2 - a^2 - b^2)^2$$
which can be written as
$$- a^4 - b^4 - c^4 + 2a^2b^2 + 2b^2c^2 + 2c^2a^2 = 0$$
or
$$(a + b + c)(a + b - c)(b + c - a)(c + a - b) = 0.$$
We are done.
