$\alpha$ and $\beta$ do not differ by an even multiple of $\pi$. If $\theta$ satisfies $$\frac{\cos\alpha}{\cos\theta}+ \frac{\sin\alpha}{\sin\theta}=\frac{\cos\beta}{\cos\theta}+\frac{\sin\beta}{\sin\theta}=1$$ then show that $$\frac{\cos\alpha\cos\beta}{\cos^2\theta}+\frac{\sin\alpha\sin\beta}{\sin^2\theta}+1=0$$
I wish to solve this problem using some elegant method, preferably complex numbers.
I've tried using the fact that $\alpha$ and $\beta$ satisfy an equation of the form $\cos x/\cos\theta + \sin x/\sin\theta = 1$, and got the required result. See my solution here: https://www.pdf-archive.com/2017/07/01/solution
I'm guessing there's an easier way to go about it. Thanks in advance!
Answer
Well, I'm not sure I can do that in a very elegant way, but it might be shorter. I'm using addition theorems, and the identity $$\sin x -\sin y=2\sin\frac{x-y}{2}\,\cos\frac{x+y}{2}$$ following immediately from them.
Multiplying the given equations by $\sin\theta\,\cos\theta,$ we get
$$\sin(\alpha+\theta)=\sin\theta\,\cos\theta=\sin(\beta+\theta),$$
but $$0=\sin(\alpha+\theta)-\sin(\beta+\theta)=2\sin\frac{\alpha-\beta}{2}\,\cos\left(\frac{\alpha+\beta}{2}+\theta\right).$$ The first factor is $\neq0$ by assumption, so $$\cos\left(\frac{\alpha+\beta}{2}+\theta\right)=0.$$ Multiplying by $2\sin\left(\frac{\alpha+\beta}{2}-\theta\right)$ and using the above identity, you get $\sin(\alpha+\beta)-\sin2\theta=0$, and this means (using $\sin2\theta=2\sin\theta\,\cos\theta$ and dividing by $\sin\theta\,\cos\theta$)
$$\frac{\sin\alpha\,\cos\beta}{\sin\theta\,\cos\theta}+\frac{\sin\beta\,\cos\alpha}{\sin\theta\,\cos\theta}=2.$$ Now you have
$$\left(\frac{\cos\alpha}{\cos\theta}+ \frac{\sin\alpha}{\sin\theta}\right)\,\left(\frac{\cos\beta}{\cos\theta}+\frac{\sin\beta}{\sin\theta}\right)-\left(\frac{\sin\alpha\,\cos\beta}{\sin\theta\,\cos\theta}+\frac{\sin\beta\,\cos\alpha}{\sin\theta\,\cos\theta}\right)=1\cdot1-2=-1,$$ and that gives your required result after simplifying.
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