If $$\cos \alpha + \cos \beta = 1$$ ,then the maximum value of $$\sin \alpha − \sin \beta$$ is

Given,

$$\cos \alpha + \cos \beta = 1$$

Squaring both sides, $$\left(\cos\ \alpha\ +\cos\ \beta\ \right)^2=1$$

or, $$\cos\ ^2\alpha\ +\cos\ ^2\beta\ +2\cos\ \alpha\ \cdot\cos\ \beta\ =1$$

or, $$1-\sin\ ^2\alpha\ +1-\sin\ ^2\beta\ +2\cos\ \alpha\ \cdot\cos\ \beta\ =1$$

or, $$1+2\cos\ \alpha\ \cdot\cos\ \beta\ =\sin^2\alpha\ +\sin^2\beta\ $$

Adding, $$-2\sin\alpha\ \sin\beta\ $$ on both sides,

$$1+2\cos\ \alpha\ \cdot\cos\ \beta-2\sin\alpha\ \sin\beta\ \ =\sin^2\alpha\ +\sin^2\beta\ -2\sin\ \alpha\ \sin\ \beta\ $$

 or, $$1+2\cos\ \left(\alpha+\beta\ \right)\ \ \ =\left(\sin\alpha\ -\sin\beta\ \right)^2$$

Now, for maximum value of $$\left(\sin\alpha−\sin\beta\right)$$, $$\cos\left(\alpha\ +\beta\ \right)$$ should be 1 (its maximum value)

So,$$\left(\sin\alpha\ -\sin\beta\right)\ ^2=1+2\times\ 1=3$$

or, $$\left(\sin\alpha\ -\sin\beta\right)=\sqrt{\ 3}$$

So, maximum value of $$\sin \alpha − \sin \beta$$ is $$\sqrt{\ 3}$$