In a triangle ABC, a[b cos C - c cos B] =

Question

In a triangle ABC, a[b cos C - c cos B] = (A) a2 (B) b2 (C) 0 (D) b2-c2

Tardigrade Admin · Accepted Answer

a[b cos C-c cos B] =(b cos C+c cos B)(b cos C-c cos B) =b2 cos 2 C-c2 cos 2 B =b2(1- sin 2 C)-c2(1- sin 2 B) =b2(1-(c2/4 R2))-c2(1-(b2/4 R2)) (∵ sin C=(C/2 R)) =b2-(b2 c2/4 R2)-c2+(c2 b2/4 R2) =b2-c2

Q. In a triangle $A BC, a [b cos C - c cos B]$ =

$a^{2}$

$b^{2}$

$0$

$b^{2} - c^{2}$

Q. In a triangle ABC,a[bcosC−ccosB] =

a2

b2

0

b2−c2

a[bcosC−ccosB] =(bcosC+ccosB)(bcosC−ccosB) =b2cos2C−c2cos2B =b2(1−sin2C)−c2(1−sin2B) =b2(1−4R2c2​)−c2(1−4R2b2​) (∵sinC=2RC​) =b2−4R2b2c2​−c2+4R2c2b2​ =b2−c2

Q. In a triangle $A BC, a [b cos C - c cos B]$ =

$a^{2}$

$b^{2}$

$0$

$b^{2} - c^{2}$