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lf `sin alpha` and `cos alpha` are the roots of the equation `ax^2 + bx + c = 0` then prove that `a^2+2ac = b^2`

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Since `sinthetaandcostheta`are the roots of equation `ax^(2)-bx+c=0`.
`:.` Sum of roots , `sintheta+costheta=-((-b))/(a)=(b)/(a)` ...(1)
and product of roots , `sinthetacostheta=(c)/(a)` ...(2)
On squaring equation (1), both sides , we get
`(sintheta+costheta)^(2)=(b^(2))/(a^(2))`
`sin ^(2)theta+cos^(2)theta+2sinthetacostheta=(b^(2))/(a^(2))`
`1+2sinthetacostheta=(b^(2))/(a^(2))` (`:' sin^(2)theta+cos^(2)theta=1)`
`1+2*(c)/(a)=(b^(2))/(a^(2))`[from(2)]
`a^(2)+2ac=b^(2)`
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