If $$\frac{\cos \alpha}{\sin \alpha + \cos \beta} + \frac{\cos \beta}{\sin \beta - \cos \alpha} = \frac{x}{\sin \alpha - \cos \beta} + \frac{\cos \beta}{\sin \beta + \cos \alpha}$$ then '$$x$$' is equal to:

$$\frac{\cos\alpha}{\sin\alpha+\cos\beta}\ -\frac{x}{\sin\alpha-\cos\beta}\ =\frac{\cos\beta}{\sin\beta+\cos\alpha}-\frac{\cos\beta}{\sin\beta-\cos\alpha}$$

$$\frac{\cos\alpha\left(\sin\alpha-\cos\beta\right)-x\left(\sin\alpha+\cos\beta\right)}{\sin^2\alpha-\cos^2\beta}=\frac{\cos\beta\left(\sin\beta-\cos\alpha\right)-\cos\beta\left(\sin\beta+\cos\alpha\right)}{\sin^2\beta-\cos^2\alpha}$$

$$\frac{\cos\alpha\left(\sin\alpha-\cos\beta\right)-x\left(\sin\alpha+\cos\beta\right)}{\sin^2\alpha-\cos^2\beta}=\frac{\cos\beta\left(\sin\beta-\cos\alpha\right)-\cos\beta\left(\sin\beta+\cos\alpha\right)}{1-\cos^2\beta-\left(1-\sin^2\alpha\right)}$$

$$\frac{\cos\alpha\left(\sin\alpha-\cos\beta\right)-x\left(\sin\alpha+\cos\beta\right)}{\sin^2\alpha-\cos^2\beta}=\frac{\cos\beta\left(\sin\beta-\cos\alpha\right)-\cos\beta\left(\sin\beta+\cos\alpha\right)}{1-\cos^2\beta-\left(1-\sin^2\alpha\right)}$$

$$\frac{\cos\alpha\left(\sin\alpha-\cos\beta\right)-x\left(\sin\alpha+\cos\beta\right)}{\sin^2\alpha-\cos^2\beta}=\frac{\cos\beta\left(\sin\beta-\cos\alpha\right)-\cos\beta\left(\sin\beta+\cos\alpha\right)}{\sin^2\alpha-\cos^2\beta}$$

$$\cos\alpha\sin\alpha-\cos\alpha\cos\beta-x\sin\alpha\ -x\cos\beta=\cos\beta\sin\beta-\cos\beta\cos\alpha-\cos\beta\sin\beta-\cos\beta\cos\alpha$$

$$\cos\alpha\sin\alpha-x\sin\alpha\ -x\cos\beta=-\cos\beta\cos\alpha$$

$$\cos\alpha\sin\alpha\ +\cos\beta\cos\alpha=x\sin\alpha\ +x\cos\beta$$

$$\cos\alpha\left(\sin\alpha\ +\cos\beta\right)=x\left(\sin\alpha\ +\cos\beta\right)$$

$$x = \cos\alpha$$