If #2tanA=3tanB# then prove that: #cos2A=(13cos2B-5)/(13-5cos2B)# ?

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Answer 1 · 2018-04-29T02:35:48

Given #2tanA=3tanB#

#=>tanA=3/2tanB#

Now

#LHS=cos2A#

#=(1-tan^2A)/(1+tan^2A)#

#=(1-9/4tan^2B)/(1+9/4tan^2B)#

#=(1-9/4*sin^2B/cos^2B)/(1+9/4*sin^2B/cos^2B)#

#=(4cos^2B-9sin^2B)/(4cos^2B+9sin^2B)#

#=(8cos^2B-18sin^2B)/(8cos^2B+18sin^2B)#

#=(4(1+cos2B)-9(1-cos2B))/(4(1+cos2B+9(1-cos2B))#

#=(4+4cos2B-9+9cos2B)/(4+4cos2B+9-9cos2B)#

#=(13cos2B-5)/(13-5cos2B)=RHS#

Proved