Question

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Answer 1

Explanation:

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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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