Question Number 113848 by deepraj123 last updated on 15/Sep/20
Answered by 1549442205PVT last updated on 16/Sep/20
![We have Σa^2 (sin^2 B−sin^2 C)= a^2 (sin^2 B−sin^2 C )+b^2 (sin^2 C−sin^2 A) +c^2 (sin^2 A−sin^2 B) From sine theorem we have: (a/(sinA))=(b/(sinB))=(c/(sinC))=2R⇒a^2 =4R^2 sin^2 A b^2 =4R^2 sin^2 B,c^2 =4R^2 sim^2 C.Replace into above equality we get Σa^2 (sin^2 B−sin^2 C) =4R^2 [sin^2 A(sin^2 B−sin^2 C)+sin^2 B(sin^2 C−sin^2 A) +sin^2 C(sin^2 A−sin^2 B)]. (Put x=sin^2 A,y=sin^2 B,z=sin^2 C ) =4R^2 [x(y−z)+y(z−x)+z(x−y)] =4R^2 (xy−xz+yz−yx+zx−zy)=0](Q113955.png)