Question Number 113848 by deepraj123 last updated on 15/Sep/20

In a △ABC, Σ a^2 (sin^2 B −sin^2 C) =

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