Question

If 2 tan a=3 tan b prove that tan(a-b)=sin2b/(5-cos2b)

Answer 1

Trigonometric Problem

Given: $2\:tanA=3\:tanB$

To prove: $tan(A-B)=\frac{sin2B}{5-cos2B}$

Step-by-step explanation:

L.H.S. $=tan(A-B)$

$=\frac{tanA-tanB}{1+tanA\:tanB}$

$=\frac{\frac{3}{2}\:tanB-tanB}{1+\frac{3}{2}tanB\:tanB}$

$=\frac{\frac{1}{2}\:tanB}{1+\frac{3}{2}\:tan^{2}B}$

$=\frac{tanB}{2+3\:tan^{2}B}$

$=\frac{2\:tanB}{4+6\:tan^{2}B}$

$=\frac{2\:tanB}{5+5tan^{2}B-1+tan^{2}B}$

$=\frac{2\:tanB}{5(1+tan^{2}B)-(1-tan^{2}B)}$

$=\frac{\frac{2\:tanB}{1+tan^{2}B}}{\frac{5(1+tan^{2}B)-(1-tan^{2}B)}{1+tan^{2}B}}$

$=\frac{\frac{2\:tanB}{1+tan^{2}B}}{5-\frac{1-tan^{2}B}{1+tan^{2}B}}$

$=\frac{sin2B}{5-cos2B}$

$=$ R. H. S.

Hence proved.