English: So, let’s starts with 1^(st) equation, you can write c ₂ in terms of c ₁ as Now substitute above equation in equation (1) we get So, this implies you have $\frac{c_{1}}{\cos\ (\mu a)\ - cos\ (\mu b)\ }\left\lbrack \left( \cos\ (\mu a)\ - cos\ (\mu b)\ \right)^{2} + \left( \sin\ (\mu a)\ - sin\ (\mu b)\ \right)^{2} \right\rbrack = 0$ For all the values of $\mu = \frac{2n\pi}{b - a}$; [C:\Users\user\Pictures\vlcsnap-2018-02-14-12h17m14s66.png] So, keeping this in mind previous equation will become $\Rightarrow \frac{c_{1}}{\cos\ (\mu a)\ - cos\ (\mu b)\ }\left\lbrack 2 - 2\left( \cos\ (\mu a).cos\ (\mu b)\ + sin\ (\mu a).sin\ (\mu b)\ \right) \right\rbrack = 0$ $\Rightarrow \frac{2c_{1}}{\cos\ (\mu a)\ - cos\ (\mu b)\ }\left\lbrack 1 - \left( \cos\ \left( \mu(b - a \right)\ \right) \right\rbrack = 0$ For $\mu = \frac{2n\pi}{b - a}$ satisfying this equation [C:\Users\user\Pictures\vlcsnap-2018-02-14-12h19m22s43.png] That implies This implies c₁ is arbitrary for which this relation is true.
Telugu: కాబట్టి, 1^(st) ఈక్వేషన్ తో ప్రారంభిద్దాం, మీరు c₂  పరంగా  c₁ ని వ్రాయవచ్చు From equation (2) ఈక్వేషన్ (2) నుంచి $c_{2} = c_{1}\frac{\sin\sin\ (\mu a)\ - sin(\mu b)}{\cos\cos\ (\mu a)\ - cos(\mu b)}$ ఇప్పుడు పై ఈక్వేషన్ ని ఈక్వేషన్ (1) లో ప్రత్యామ్నాయం చేయండి.