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If A, B, C are the angles of a triangle ...

If A, B, C are the angles of a triangle then `sin^(2)A+sin^(2)B+sin^(2)C-2cosAcosBcosC` is equal to

A

1

B

2

C

3

D

4

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Verified by Experts

The correct Answer is:
B
`sin^(2)A+sin^(2)B+sin^(2)C`
`=1-cos^(2)A+1-cos^(2)B+sin^(2)C`
`=2-cos^(2)A-cos(B+C)cos(B-C)`
`=2-cosA[cosA-cos(B-C)]`
`=2-cosA[-cos(B+C)-cos(B-C)]`
`=2+cosA.2cosBcosC`
`therefore sin^(2)A+sin^(2)B+sin^(2)C-2cosAcosBcosC=2`
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