Question

a²(cos²B-cos²C) + b²(cos²C-cos²A) + c²(cos²A-cos²B) =0​

Answer 1

a^2(cos^2b-cos^2c)+b^2(cos^c-cos^2a)+c^2(cos^2a-cos^2b)=0


a²(cos2B-cos2C)+b²(cos2C-cos2A)+c²(cos2A-cos2B)=0
a²(cos2B-cos2C)+b²(cos2C-cos2A)+c²(cos2A-cos2B)
=a²(1-2sin²B-1+2sin²C)+b²(1-2sin²C-1+2sin²A)+c²(1-2sin²A-1+2sin²B)
=a²(2sin²C-2sin²B)+b²(2sin²A-2sin²C)+c²(2sin²B-2sin²A)
=2a²(sin²C-sin²B)+2b²(sin²A-sin²C)+2c²(sin²B-sin²A)
=2a²(c²-b²)/(4R²)+2b²(a²-c²)/(4R²)+2c²(b²-a²)/(4R²)
=2/(4R²)·[a²(c²-b²)+b²(a²-c²)+c²(b²-a²)]
=2/(4R²)·(a²c²-a²b²+a²b²-b²c²+b²c²-a²c²)
=2/(4R²)·0
=0

Mark it as brainly....

Answer 2

$\large\bf{\underline\green{❥solution:-}}$

a^2(cos^2b-cos^2c)+b^2(cos^c-cos^2a

)+c^2(cos^2a-cos^2b)=0

a²(cos2B-cos2C)+b²(cos2C-cos2A)+c²(cos2A-cos2B)=0

a²(cos2B-cos2C)+b²(cos2C-cos2A)+c²(cos2A-cos2B)

=a²(1-2sin²B-1+2sin²C)+b²(1-2sin²C-1+2sin²A)+c²(1-2sin²A-1+2sin²B)

=a²(2sin²C-2sin²B)+b²(2sin²A-2sin²C)+c²(2sin²B-2sin²A)

=2a²(sin²C-sin²B)+2b²(sin²A-sin²C)+2c²(sin²B-sin²A)

=2a²(c²-b²)/(4R²)+2b²(a²-c²)/(4R²)+2c²(b²-a²)/(4R²)

=2/(4R²)·[a²(c²-b²)+b²(a²-c²)+c²(b²-a²)]

=2/(4R²)·(a²c²-a²b²+a²b²-b²c²+b²c²-a²c²)

=2/(4R²)·0

=0