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

1 Answer
Apr 29, 2018

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