If sin theta and cos theta are roots of equation ax^2+bx+c=0 prove that a^2-b^2+2ac=0
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Let me assume theta is a
Sina+cosa=-b/a
Sina*cosa=c/a
By applying (A+B)^2 formulae
By A is sina, B is cosa
(Sina+cosa)^2=sin^2a+cos^2a +2sinacosa
b^2/a^2=1+2ca
Thus by calculation we get
a^2-b^2+2ac=0
..hence proved..
Sina+cosa=-b/a
Sina*cosa=c/a
By applying (A+B)^2 formulae
By A is sina, B is cosa
(Sina+cosa)^2=sin^2a+cos^2a +2sinacosa
b^2/a^2=1+2ca
Thus by calculation we get
a^2-b^2+2ac=0
..hence proved..
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