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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Trignometry
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If $\sin\theta=n\sin(\theta+2\alpha)$ then $tan(\theta+\alpha)$ is equal to

$\begin{array}{1 1}(a)\;\large\frac{1+n}{2-n}\normalsize\tan\alpha&(b)\large\frac{1-n}{1+n}\normalsize\tan\alpha\\(c)\;\tan\alpha&(d)\;\large\frac{1+n}{1-n}\normalsize\tan\alpha\end{array}$

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1 Answer

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We have
$\sin\theta=n\sin(\theta+2\alpha)$
$\Rightarrow \large\frac{\sin(\theta+2\alpha)}{\sin\theta}=\frac{1}{n}$
$\Rightarrow \large\frac{\sin(\theta+2\alpha)+\sin\theta}{\sin(\theta+2\alpha)-\sin\theta}$
$\Rightarrow \large\frac{1+n}{1-n}$
$\Rightarrow \large\frac{2\sin(\theta+\alpha).\cos\alpha}{2\sin\alpha\cos(\theta+\alpha)}=\frac{1+n}{1-n}$
$\tan(\theta+\alpha)=\large\frac{1+n}{1-n}$$\tan\alpha$
Hence (d) is the correct option.
answered Oct 14, 2013 by sreemathi.v
 

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