If in a triangle ABC, sin⁡A:sin⁡C=sin⁡(A−B):sin⁡(B−C)  then, a2:b2:c2 are in 

If in a triangle ABC, sinA:sinC=sin(AB):sin(BC)  then, a2:b2:c2 are in 

  1. A

    A.P.

  2. B

    G.P.

  3. C

    H.P.

  4. D

    none of these 

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    Solution:

    we have, 

    tan8212=cot712=cos712sin712=2cos27122sin712cos712=1+cos15sin15=1+3+12231=22+3+131=(22+3+1)(3+1)(3+1)(31)=26+22+3+3+3+12=6+2+4+3=2+3+4+6

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