# Sin A and cosA are the zeros of polynomial ax2 +bx+c then prove that a2+2ac=b2

Solution:

Given equation

ax^2 + bx + c = 0,Root of equation is SinA &CosA

Now

SinA + CosA = -b/a

SinA*CosA= c/a

We know Sin^2A + Cos^2A=1

( SinA + CosA)^2= Sin^2A +Cos^2A+2SinA*CosA

( -b/a )^2 =1+2c/a

b^2/a^2 =1 + 2c/a

a^2 + 2ac =b^2

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