In a △ABC,tan⁡A2=56,tan⁡C2=25,then

In a ABC,tanA2=56,tanC2=25,then

  1. A

    a, c, b are in A.P.

  2. B

    a, b,c are in A.P.

  3. C

    b, a, c are in AP.

  4. D

    a,b,c are in G.P.

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    Solution:

    We have,

    tanA2=56 and, tanC2=25tanA2tanC2=13(sb)(sc)s(sa)(sb)(sa)s(sc)=13sbs=133s3b=s

     2s=3b2b=a+ca,b,c are in  A.P.

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