If `sinalpha+sinbeta=a` and `cosalpha+cosbeta=b` prove that: `cos(alpha-beta)=((a^2+b^2-2)/2)`

Explanation

The given equations are `sin alpha+sin beta=a and cos alpha+cos beta=b`

On squaring `sin alpha+sin beta=a and cos alpha+cos beta=b` and adding them, we get `sin ^{2} `
`alpha+sin ^{2} beta+2 times sin alpha sin beta+cos ^{2} alpha+cos ^{2} beta+2 times cos alpha cos beta=a^{2}+b^{2}`
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Rightarrow 1+1+2(sin alpha sin beta+cos alpha cos beta)=a^{2}+b^{2}`
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Rightarrow 2(sin alpha sin beta+cos alpha cos beta)=a^{2}+b^{2}-2`
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Rightarrow 2 cos (alpha-beta)=a^{2}+b^{2}-2(because cos (A-B)=sin A sin B+cos A cos B) `
`Rightarrow cos (alpha-beta)=frac{a^{2}+b^{2}-2}{2}`