<s>[INST] If $\alpha$, $\beta$ are two values of $\theta$ satisfying the equation $\cos\theta/a+\sin\theta/b=1/c$, prove that $\cot ((\alpha+\beta)/2) = b/a$ What I did was 
$$b\ \cos (\theta) + a \sin (\theta) = \dfrac{ab}{c} \\
b\ \cos (\theta) = \frac{ab}{c} - a\ \sin (\theta) $$
Square both sides and using sum of roots and product of roots as
$$\alpha + \beta = \dfrac{2a^2b}{c(a^2+b^2)}\ \  \text{and}\ \ 
  \alpha\beta = \dfrac{a^2b^2 - b^2c^2}{c^2(a^2+b^2)}$$
Now 
$$b\ \cos (\theta) - \dfrac{ab}{c} = -a\ \sin (\theta)$$
Square both sides and using sum of roots and product of roots as
$$ \alpha + \beta = \dfrac{2ab^2}{c(a^2+b^2)} \\
  \alpha\beta = \dfrac{a^2b^2 - a^2c^2}{c^2(a^2+b^2)} $$
I don't know how to solve further.
 [/INST] Hint.  Write
$$\theta=\frac{\alpha+\beta}{2}\ ,\quad \phi=\frac{\alpha-\beta}{2}\ .$$
You need to find $\cot\theta$, and you are given that
$$\frac{\cos(\theta+\phi)}{a}+\frac{\sin(\theta+\phi)}{b}
  =\frac{\cos(\theta-\phi)}{a}+\frac{\sin(\theta-\phi)}{b}\ .$$
See if you can take it from here.
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