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

If 2tanA=3tanB2tanA=3tanB then prove that: cos2A=(13cos2B-5)/(13-5cos2B)cos2A=(13cos2B-5)/(13-5cos2B) ?