If $2 tan A = 3 tan B$ $2tanA=3tanB$ then prove that: $cos 2 A = \frac{13 cos 2 B - 5}{13 - 5 cos 2 B}$ $cos2A=(13cos2B-5)/(13-5cos2B)$ ?

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If $2 tan A = 3 tan B$ $2tanA=3tanB$ then prove that: $cos 2 A = \frac{13 cos 2 B - 5}{13 - 5 cos 2 B}$ $cos2A=(13cos2B-5)/(13-5cos2B)$ ?

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Answer 1 · 2018-04-29T02:35:48

Given $2 tan A = 3 tan B$ $2tanA=3tanB$

$\Rightarrow tan A = \frac{3}{2} tan B$ $=>tanA=3/2tanB$

Now

$L H S = cos 2 A$ $LHS=cos2A$

$= \frac{1 - {tan}^{2} A}{1 + {tan}^{2} A}$ $=(1-tan^2A)/(1+tan^2A)$

$= \frac{1 - \frac{9}{4} {tan}^{2} B}{1 + \frac{9}{4} {tan}^{2} B}$ $=(1-9/4tan^2B)/(1+9/4tan^2B)$

$= \frac{1 - \frac{9}{4} \cdot \frac{{sin}^{2} B}{{cos}^{2} B}}{1 + \frac{9}{4} \cdot \frac{{sin}^{2} B}{{cos}^{2} B}}$ $=(1-9/4*sin^2B/cos^2B)/(1+9/4*sin^2B/cos^2B)$

$= \frac{4 {cos}^{2} B - 9 {sin}^{2} B}{4 {cos}^{2} B + 9 {sin}^{2} B}$ $=(4cos^2B-9sin^2B)/(4cos^2B+9sin^2B)$

$= \frac{8 {cos}^{2} B - 18 {sin}^{2} B}{8 {cos}^{2} B + 18 {sin}^{2} B}$ $=(8cos^2B-18sin^2B)/(8cos^2B+18sin^2B)$

$= \frac{4 (1 + cos 2 B) - 9 (1 - cos 2 B)}{4 (1 + cos 2 B + 9 (1 - cos 2 B))}$ $=(4(1+cos2B)-9(1-cos2B))/(4(1+cos2B+9(1-cos2B))$

$= \frac{4 + 4 cos 2 B - 9 + 9 cos 2 B}{4 + 4 cos 2 B + 9 - 9 cos 2 B}$ $=(4+4cos2B-9+9cos2B)/(4+4cos2B+9-9cos2B)$

$= \frac{13 cos 2 B - 5}{13 - 5 cos 2 B} = R H S$ $=(13cos2B-5)/(13-5cos2B)=RHS$

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If $2 tan A = 3 tan B$ $2tanA=3tanB$ then prove that: $cos 2 A = \frac{13 cos 2 B - 5}{13 - 5 cos 2 B}$ $cos2A=(13cos2B-5)/(13-5cos2B)$ ?

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If 2tanA=3tanB2tanA=3tanB then prove that: cos2A=13cos2B−513−5cos2Bcos2A=(13cos2B-5)/(13-5cos2B) ?

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If $2 tan A = 3 tan B$ $2tanA=3tanB$ then prove that: $cos 2 A = \frac{13 cos 2 B - 5}{13 - 5 cos 2 B}$ $cos2A=(13cos2B-5)/(13-5cos2B)$ ?