MCQ
If $\tan\alpha=\frac{1-\cos\beta}{\sin\beta},$ then:
  • A
    $\tan3\alpha=\tan2\beta$
  • B
    $\tan2\alpha=\tan\beta$
  • C
    $\tan2\beta=\tan\alpha$
  • D
    None of these

Answer

  1. $\tan2\alpha=\tan\beta$

Solution:

$\tan\alpha=\frac{1-\cos\beta}{\sin\beta}$

$=\frac{2\sin^2\frac{\beta}{2}}{2\sin\frac{\beta}{2}\cos\frac{\beta}{2}}$

$=\frac{\sin\frac{\beta}{2}}{\cos\frac{\beta}{2}}$

$\Rightarrow\tan\alpha=\tan\frac{\beta}{2}$

$\Rightarrow\alpha=\frac{\beta}{2}$

$\Rightarrow2\alpha=\beta$

$\therefore\tan2\alpha=\tan\beta$

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