if alpha nad beta are the roots of acosx+bcosx=c show
that cos(alpha+beta)=(a²-b²) / (a²+b²)

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Solution

$a \cos x + b \sin x = c N o w s o l v i n g t h i s e q u a t i o n b y t a k i n g x = θ f o r s i m p l i c i t y, a (\frac{1 - \tan^{2} \frac{θ}{2}}{1 + \tan^{2} \frac{θ}{2}}) + b (\frac{2 \tan θ}{1 + \tan^{2} \frac{θ}{2}}) = c [A s \cos θ = \frac{1 - \tan^{2} \frac{θ}{2}}{1 + \tan^{2} \frac{θ}{2}} a n d \sin θ = \frac{2 \tan θ}{1 + \tan^{2} \frac{θ}{2}}] O r (a + c) \tan^{2} \frac{θ}{2} - 2 b \tan \frac{θ}{2} + (c - a) = 0 A s i t i s q u a d r a t i c i n \tan \frac{θ}{2} L e t i t s r o o t s a s \tan \frac{A}{2} a n d \tan \frac{B}{2} S o s u m o f r o o t s = \tan \frac{A}{2} + \tan \frac{B}{2} = \frac{2 b}{a + c} A n d p r o d u c t o f r o o t s = \tan \frac{A}{2} \times \tan \frac{B}{2} = \frac{c - a}{c + a} A n d \tan (\frac{A + B}{2}) = \frac{\tan \frac{A}{2} + \tan \frac{B}{2}}{1 - \tan \frac{A}{2} \tan \frac{B}{2}} = \frac{2 b / (a + c)}{1 - [(c - a) / (c + a)]} = \frac{2 b}{2 a} = \frac{b}{a} A n d \cos (A + B) = \frac{1 - \tan^{2} (\frac{A + B}{2})}{1 + \tan^{2} (\frac{A + B}{2})} = \frac{1 - b^{2} / a^{2}}{1 + b^{2} / a^{2}} = \frac{a^{2} - b^{2}}{a^{2} + b^{2}} H e n c e p r o v e d .$
Where A= $α and B = β$

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