Given that, 1 is a root of ax2+bx+c=0 ‌⇒‌‌a+b+c=0 ‌∴‌‌E1:a+b+c=0‌ is true. ‌ Since cos‌θ,sin‌θ are the roots of ‌ax2+bx+c=0 ∴‌sin‌θ+cos‌θ=−‌
b
a
. . . (i) ‌ and ‌sin‌θ‌cos‌θ=‌‌
c
a
On squaring both sides of equation (i) ‌(sin‌θ+cos‌θ)2‌=‌
b2
a2
⇒sin‌2θ+cos2θ+2sin‌θ‌cos‌θ‌=‌
b2
a2
⇒1+2(‌
c
a
)‌=‌
b2
a2
⇒2⋅‌
c
a
‌=‌
b2−a2
a2
⇒‌−a2+b2‌=2ac ‌∴‌E2:b2−a2‌=2ac‌ is true ‌ ⇒E1 and E2 both are true.