4R2
[ – sin2Asin2B + sin2Asin2C – sin2Bsin2C + sin2Asin2B – sin2Asin2C + sin2Bsin2C ]
= 4R2
[0]
= 0 [Proved]
In any ΔABC, prove that
In any ΔABC, prove that
4R2
[ – sin2Asin2B + sin2Asin2C – sin2Bsin2C + sin2Asin2B – sin2Asin2C + sin2Bsin2C ]
= 4R2
[0]
= 0 [Proved]