If α and β are the solution of a cos θ + b sin θ = c, then show that cos( α + β ) = 
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we have a cos θ + b sin θ = c ...(i)
⇒ a cos θ = c – b sin θ ⇒ a 2 cos 2 θ = (c – b sin θ ) 2 ⇒ a 2 (1 – sin 2 θ ) = c 2 – 2bcsin θ + b 2 sin 2 θ
⇒ (a 2 + b 2 ) sin 2 θ – 2bc sin θ + (c 2 – a 2 ) = 0 ...(ii)
Since α , β are roots of equation (i). Therefore, sin α and sin β are roots of equation (ii)
∴ sin α sin β =
...(iii)
Again, acos θ + b sin θ = c ⇒ bsin θ = c – acos θ ⇒ b 2 sin 2 θ = (c – a cos θ ) 2
⇒ b 2 (1 – cos 2 θ ) = (c – acos θ ) 2 ⇒ (a 2 + b 2 ) cos 2 θ – 2ac cos θ + c 2 – b 2 = 0 ...(iv)
It is given that α , β are the the roots of equation (i), So, cos α , cos β are the roots of equation (iv).
∴ cos α cos β =
...(v)
Now,cos( α + β ) = cos α cos β – sin α sin β⇒ cos( α + β ) =
–
= 
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