यदि `A+B+C =pi,` सिद्ध कीजिए कि
`sin^(2)A+sin^(2)B+sin^(2)C=2[1+cos A cos B cos C]`

Explanation

बायाँ पक्ष `=sin^(2)A+sin^(2)B+sin^(2)C`
`=(1)/(2)(1-cos2A)+(1)/(2)(1-cos 2B)+(1)/(2)(1-cos 2C)`
`=(3)/(2)-(1)/(2)(cos2A+cos2B+cos2C)`
`=(3)/(2)-(1)/(2)[2cos(A+B)cos(A-B)+2cos^(2)C-1]`
`=(3)/(2)-(1)/(2)[2cos (pi-C)cos(A-B)+2cos^(2)C-1]`
`=(3)/(2)-(1)/(2)[-2 cos C cos (A-B)+2cos^(2)C-1]`
`=(3)/(2)-(1)/(2)[-2cos C{cos (A-B)-cos C}-1]`
`=(3)/(2)-(1)/(2)[-2cos C{cos(A-B)-cosC}-1]`
`=(3)/(2)-(1)/(2)[-1-2 cos C{cos(A-B)-cos{pi-(A+B)}]`
`=(3)/(2)-(1)/(2)[-1-2cosC{cos(A-B)+cos (A+B)}]`
`=(3)/(2)-(1)/(2)[-1-2cos C(2cos AcosB)]`
`=(3)/(2)-(1)/(2)(-1-4cosAcos BcosC)`
`=2+2 cos A cos B cosC`
`=2(1+cos A cos B cos C)=` दायाँ पक्ष