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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.

gebruiker
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Shubham
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  • Related (but not duplicate): "Cosine of the sum of two solutions of trigonometric equation $a\cos\theta+b\sin\theta=c$". Now, my answer to that question involves a diagram that clearly shows (in the notation used there) $\cot((\theta+\phi)/2) = a/b$. Replacing that question's $a$ and $b$ (and $c$) with your question's $1/a$ and $1/b$ (and $1/c$) gives the result you seek. Other answers to that question provide additional approaches. – Blue Sep 18 '14 at 01:49
  • @Shubham I made an [edit] to your question, in which I deleted the part where you answered the question. The question box is for questions, the answer box is for answers. I would recomend that you self-answer your question in cases like this (i.e. post an answer to your own question). You could still do that now. (If you want to read back what you wrote, you can check the revision history.) Be shure to use MathJaX when you do. I would also recommend that in stead of posting a "thank-you comment", you upvote and/or accept useful answers. – gebruiker Feb 24 '16 at 09:31

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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.

David
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