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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

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

Grade:11

1 Answers

Yash Baheti IIT Roorkee
askIITians Faculty 97 Points
9 years ago
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.

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