If α and β are the roots of acosx + bsinx = c then show that cos( α – β ) = (2c^2-(a^2+b^2))/a^2+b^2

GIVEN EQUATION : acosx + bsinx = c 
Provided roots of the equation - α and β 

HENCE , acosα + bsinα = c , acosβ + bsinβ = c 

Subtracting the subsequent second equation from the first , we obtain ; 

a(cosα - cosβ) + b(sinα - sinβ) = 0 
or , 
b ( sinα - sinβ ) - a ( cosβ - cosα ) = 0 
or , 
2b cos α + β/2 sin α - β/2 = 2asin α + β/2 sin α- β /2
or , 
Tan α+ β /2 =b/a [ because α , β are different angles , hence can be substituted as sinα-β/2≠0 

Hence , Sin (α+β)= 2Tan α+β/2 / 1 + Tan² α+β /2 
= 2 { b/a} / 1 + b²/a² = 2ab / a² + b² 

Hence , by further reductions , 

Tan (α+β)=2Tan(α+β/2)/1-Tan²( α+β /2 ) = 2ab / a² - b²