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.
 A: From equation $$a\sin(\alpha)+b\sin(\beta)+c\sin(\gamma)=0$$
$$a\sec(\alpha)+b\sec(\beta)+c\sec(\gamma)=0$$
using cross-multiplication:
$$\frac{a}{\frac{\sin\beta}{\cos\gamma}-\frac{\sin\gamma}{\cos\beta}}=\frac{b}{\frac{\sin\gamma}{\cos\alpha}-\frac{\sin\alpha}{\cos\gamma}}=\frac{c}{\frac{\sin\alpha}{\cos\beta}-\frac{\sin\beta}{\cos\alpha}}=k$$
Using this and putting the value in equation $a\cos(\alpha)+b\cos(\beta)+c\cos(\gamma)=0$ ,
$$\cos\alpha \left(\frac{\sin\beta}{\cos\gamma}-\frac{\sin\gamma}{\cos\beta}\right)+\cos\beta \left(\frac{\sin\gamma}{\cos\alpha}-\frac{\sin\alpha}{\cos\alpha}\right)+\cos\gamma \left(\frac{\sin\alpha}{\cos\beta}-\frac{\sin\beta}{\cos\alpha}\right)=0$$
Which simplifies to,
$$\cos^2\alpha(\sin2\beta-\sin2\gamma)+\cos^2\beta(\sin2\gamma-\sin2\alpha)+\cos^2\gamma(\sin2\alpha-\sin2\beta)=0$$
$$(\cos2\alpha+1)(\sin2\beta-\sin2\gamma)+(\cos2\beta+1)(\sin2\gamma-\sin2\alpha)+(\cos2\gamma+1)(\sin2\alpha-\sin2\beta)=0$$
Which simplifies to,
$$\sin(\alpha-\beta) \sin(\beta-\gamma) \sin(\gamma-\alpha)=0~~~~~~~~~~~~~~~~(i)$$
In equations $1$ and $2$, shift the $\cos$ terms to other side and square both sides ,
adding them we get,
$$a^2+b^2+2ab\cos(\alpha-\beta)=c^2$$
Which gives the value of
$$\cos(\alpha-\beta)=\frac{c^2-a^2-b^2}{2ab}$$
From this we get the value of $\sin(\alpha-\beta)$ by using $\sin^2x+\cos^2x=1$
Similarly we can get to $\sin(\beta-\gamma)$ and $\sin(\gamma-\alpha)$
Putting them in equation (i) and squaring, we get
$$a^4+b^4+c^4-2b^2c^2-2a^2b^2-2c^2a^2=0$$
A: 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.
