MCQ
If $2\tan\alpha=3\tan\beta$ then $\tan(\alpha-\beta)=$
  • $\frac{\sin2\beta}{5-\cos2\beta}$
  • B
    $\frac{\cos2\beta}{5-\cos2\beta}$
  • C
    $\frac{\sin2\beta}{5+\cos2\beta}$
  • D
    None of these

Answer

Correct option: A.
$\frac{\sin2\beta}{5-\cos2\beta}$
Given:
$2\tan\alpha=3\tan\beta$
Now,
$\tan(\alpha+\beta)=\frac{\tan\alpha-\tan\beta}{1+\tan\alpha\tan\beta}$
$=\frac{\frac{3}{2}\tan\beta-\tan\beta}{1+\Big(\frac{3}{2}\tan\beta\Big)\tan\beta}$
$=\frac{3\tan\beta-2\tan\beta}{2+3\tan^2\beta}$
$=\frac{\tan\beta}{2+3\tan^2\beta}$
$=\frac{\frac{\sin\beta}{\cos\beta}}{2+3\frac{\sin^2\beta}{\cos^2\beta}}$
$=\frac{\sin\beta\cos\beta}{2\cos\beta+3\sin\beta}$
$=\frac{\sin\beta\cos\beta}{2+\sin^2\beta}$
$=\frac{2\sin\beta\cos\beta}{4+2-\cos2\beta}$
$=\frac{\sin2\beta}{4+1-cos2\beta}$
$\therefore\tan(\alpha+\beta)=\frac{\sin2\beta}{5-\cos2\beta}$

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