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MATHS
In any triangle ABC, prove that
a
2
b
2
−c
2
sin2A+
b
2
c
2
−a
2
sin2B+
c
2
a
2
−b
2
sin2C=0
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ANSWER
Let,
sinA
a
=
sinB
b
=
sinC
c
=k
⟹a=ksinA, b=ksinB, c=ksinC
Now,
LHS =
a
2
(b
2
−c
2
)
sin2A+
b
2
(c
2
−a
2
)
sin2B+
c
2
(a
2
−b
2
)
sin2C
LHS =
a
2
(b
2
−c
2
)
2sinAcosA+
b
2
(c
2
−a
2
)
2sinBcosB+
c
2
(a
2
−b
2
)
2sinCcosC
LHS =
a
2
(b
2
−c
2
)
(
k
2a
)(
2bc
b
2
+c
2
−a
2
)+
b
2
(c
2
−a
2
)
(
k
2b
)(
2ca
a
2
+c
2
−b
2
)+
c
2
(a
2
−b
2
)
(
k
2c
)(
2ba
b
2
−c
2
+a
2
)
LHS =
kabc
1
[(b
2
−c
2
) (b
2
+c
2
−a
2
)+(c
2
−a
2
) (c
2
+a
2
−b
2
)+(a
2
−b
2
) (a
2
+b
2
−c
2
)]
LHS =
kabc
1
[0]
LHS = 0
LHS = RHS
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