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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.

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  • $\begingroup$ I think the first equation does not fit the sequence. $\endgroup$ – Moti May 18 '19 at 4:02
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    $\begingroup$ Considering you tagged it with triangles, is alpha beta gamma angles of a triangle? $\endgroup$ – user541396 May 19 '19 at 3:01
  • $\begingroup$ This problem is from Trigonometry by S.L.Loney $\endgroup$ – Umesh shankar May 19 '19 at 14:53
  • $\begingroup$ @Umeshshankar I have SL loney, please give pase number, and questions number. $\endgroup$ – user541396 May 19 '19 at 15:21
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    $\begingroup$ It's hard to know what the author expects here; you might say something about what topics Loney covers near this problem. That said, a brute force approach (with complex exponentials and resultants) gives $$abc(a+b+c)(-a+b+c)(a-b+c)(a+b-c)=0$$ (some of whose factors may be extraneous) without assuming any relation among the angles. Perhaps the point is to calculate coefficient matrix' determinant, and to see that it simplifies to $$\sin(\alpha-\beta)\sin(\beta-\gamma)\sin(\gamma-\alpha)\sec\alpha\sec\beta\sec\gamma$$ If that's non-zero, then $a=b=c=0$; if it's zero, you get special cases. $\endgroup$ – Blue May 19 '19 at 17:38
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From equation $3$ and $2$, using cross-multiplication:

$$\frac{a}{\frac{\sin\beta}{\cos\gamma}-\frac{\sin\gamma}{\cos\beta}}=\frac{b}{\frac{\sin\gamma}{\cos\alpha}-\frac{\sin\alpha}{\cos\gamma}}=\frac{c}{\frac{\sin\alpha}{\cos\beta}-\frac{\sin\beta}{\cos\gamma}}=k$$

Using this and putting the value in equation $1$,

$$\cos\alpha \left(\frac{\sin\beta}{\cos\gamma}-\frac{\sin\gamma}{\cos\beta}\right)+\cos\beta \left(\frac{\sin\gamma}{\cos\alpha}-\frac{\sin\alpha}{\cos\alpha}\right)+\cos\gamma \left(\frac{\sin\alpha}{\cos\beta}-\frac{\sin\beta}{\cos\alpha}\right)=0$$

Which simplyfies 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 simplyfies to,#

$$\sin(\alpha-\beta) \sin(\beta-\gamma) \sin(\gamma-\alpha)=0$$

In equations $1$ and $2$, shift the $\cos$ terms to other side and square both sides for both the equations, 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 #, and squaring we get

$$a^4+b^4+c^4-2b^2c^2-2a^2b^2=0$$

Note:It took a lot of formatting, so if any formatting error is there please point out.

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