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Questions  

If A,B,C are the angles of a triangle and  cosB+cosC=4sin2A2, then tanB2tanC2 is equal to

a
1/2
b
1/3
c
2/3
d
3/2

detailed solution

Correct option is B

cos⁡B+cos⁡C=4sin2⁡(A/2)⇒ 2cos⁡B+C2cos⁡B−C2=4sin⁡A2sin⁡π2−B+C2⇒ 2sin⁡A2cos⁡B−C2=4sin⁡A2cos⁡B+C2⇒ cos⁡B2cos⁡C2+sin⁡B2sin⁡C2=2cos⁡B2cos⁡C2−sin⁡B2sin⁡C2⇒3sin⁡B2sin⁡C2=cos⁡B2cos⁡C2⇒tan⁡B2tan⁡C2=13.

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