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If `cos 2B =cos(A+C)/cos( A-C)`, then `tanA, tanB, tanC` are in

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If `cos 2B =cos(A+C)/cos( A-C)`, then `tanA, tanB, tanC` are in
A. AP
B. GP
C. HP
D. none of these.
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Correct Answer - B
`(cos 2B)/(1)=(cos(A+C))/(cos(A-C))`
Applying componendo and dividendo, we get
`(1-cos2B)/(1+cos2B)=(cos(A-C)-cos(A+C))/(cos(A-C)+cos(A+C))`
or `(2sin^(2)B)/(2cos^(2)B)=(2sinA sinC)/(2cosA cos C)`
Thus, `tanA,tanB,tanC` are in GP.
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