Question

If cos 2b= cos(a+c)/cos(a-c) then tan a , tan b, tan c are in

Answer 1

cos(2B) = 2 cos^2(B) - 1 = 2/sec^2(B) -1 = (1-tan^2(B))/(1+tan^2(B))

Then, clearing the fraction:

cos(A-C) - Cos(A-C)tan^2(B) = cos(A+C) + cos(A+C)tan^2(B)

tan^2(B)(cos(A+C)+cos(A-C)) = cos(A-C)-cos(A+C)

tan^2(B)*2cos(A)cos(C) = 2sin(A)sin(C)

tan^2(B) = tan(A)tan(C) proving they are in G.P

Answer 2

tan a, tan b and tan c are in G.P

Step-by-step explanation:

Given:

We have,

$cos2b = \frac{cos(a+c)}{cos(a-c)}$

⇒ $\frac{cos2b }{1}=\frac{cos(a+c)}{cos(a-c)}$

$\frac{1-cos2b}{1+cos2b} =\frac{cos(a-c)-cos(a+c)}{cos(a-c)+cos(a+c)}$   (By componendo−dividendo)

$\frac{2sin2b}{2cos2b} =\frac{2sin a sin c}{2cos a cos c}$  

$tan^2b$ = tan a tan c

∴ tan a, tan b and tan c are in G.P