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Home  >>  CBSE XI  >>  Math  >>  Trigonometric Functions
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If $m\sin \theta=n\sin(\theta+2\alpha)$,then prove that $\tan(\theta+\alpha)\cot\alpha=\large\frac{m+n}{m-n}$

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Toolbox:
  • $\sin A+\sin B=2\sin \large\frac{A+B}{2}$$.\cos \large\frac{A-B}{2}$
  • $\sin A-\sin B=2\cos \large\frac{A+B}{2}$$.\sin \large\frac{A-B}{2}$
$m\sin \theta=n\sin(\theta+2\alpha)$
$\large\frac{\sin(\theta+2\alpha)}{\sin \theta}=\frac{m}{n}$
by componendo and dividendo
$\large\frac{\sin(\theta+2\alpha)+\sin \theta}{\sin(\theta+2\alpha)-\sin \theta}=\frac{m+n}{m-n}$
$\large\frac{2\sin\Large\frac{(\theta+2\alpha+\theta)}{2}.\normalsize\cos \Large\frac{(\theta+2\alpha-\theta)}{2}}{2\cos\Large\frac{(\theta+2\alpha+\theta)}{2}.\normalsize \sin \Large\frac{(\theta+2\alpha-\theta)}{2}}=\frac{m+n}{m-n}$
$\large\frac{\sin(\theta+\alpha).\cos 2\alpha}{\cos(\theta+\alpha).\sin 2\alpha}=\frac{m+n}{m-n}$
$\tan(\theta+\alpha)\cot 2\alpha=\large\frac{m+n}{m-n}$
Hence proved.
answered Apr 18, 2014 by sreemathi.v
 

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