यदि `A+B+C=pi` हो, तो सिद्ध कीजिए कि - `sin^(2)A+sin^(2)B+sin^(2)C=2+2cos A cos B cos C`.

Explanation

`because A+B+C = pi rArr A+B = pi - C` एवं `C = pi-(A+B)`
`L.H.S. = sin^(2)A+sin^(2)A+sin^(2)B + sin^(2)C`
`=1-cos^(2)A+sin^(2)B+1-cos^(2)C`
`=2-(cos^(2)A-sin^(2)B)-cos^(2)C`
`rArr " " L.H.S. = 2-cos (A+B)cos(A-B)-cos^(2)C`
`=2-cos (pi-C)cos(A-B)-cos^(2)C`
`=2+cos C cos (A-B)-cos^(2)C`
`= 2+cos C [cos (A-B)-cos C]`
`= 2+cos C[cos(A-B)-cos {pi - (A+B)}]`
`= 2+cos C [cos (A-B)+cos (A+B)]`
`= 2+cos C.2 cos A cos B`
`= 2+2 cos A cos B cos C = R.H.S. " "` इति सिद्धम्