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If cos2B=(cos(A+C))/("cos"(A-C)), then t...

If `cos2B=(cos(A+C))/("cos"(A-C)),` then `tanA ,t a n B ,tanC` are in A.P. (b) G.P. (c) H.P. (d) none of these

A

AP

B

GP

C

HP

D

none of these.

Text Solution

Verified by Experts

The correct Answer is:
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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