IN A TRIANGLE ABC bcsin^2A/cosA+cosBcosC is equal to

(A) b^2+c^2

(B)bc

(C)a^2

(D)a^2+bc

Hi
Please follow the steps give to reach the answer :

The correct option is C
Expand the COS a of denominator by writing it as COS [pi- (B+c)].
COS B COS C will cancel out and SIN B SIN C will be remaining.
Now apply the SINE Rule. You will get the answer.