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Home  >>  CBSE XI  >>  Math  >>  Trigonometric Functions
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If $\tan\alpha=\large\frac{1}{7}$$,\tan \beta=\large\frac{1}{3}$,then $\cos 2\alpha$ is equal to

$\begin{array}{1 1}(A)\;\sin 2\beta&(B)\;\sin 4\beta\\(C)\;\sin 3\beta&(D)\;\cos 2\beta\end{array} $

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1 Answer

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Toolbox:
  • $\cos 2\alpha=\large\frac{1-\tan^2\alpha}{1+\tan^2\alpha}$
  • $\sin 2\alpha=2\sin \alpha\cos \alpha$
  • $\sin 2\alpha=\large\frac{2\tan\alpha}{1+\tan^2\alpha}$
Given :
$\tan\alpha=\large\frac{1}{7}$$,\tan \beta=\large\frac{1}{3}$
$\cos 2\alpha=\large\frac{1-\tan^2\alpha}{1+\tan^2\alpha}$
$\Rightarrow \large\frac{1-\Large\frac{1}{49}}{1+\Large\frac{1}{49}}$
$\Rightarrow \large\frac{48}{50}$
$\Rightarrow \large\frac{24}{25}$
$\sin 4\beta=2\sin 2\beta.\cos 2\beta$
$\Rightarrow 2\times \large\frac{2\tan \beta}{1+\tan^2\beta}.\frac{1-\tan^2\beta}{1+\tan^2\beta}$
$\Rightarrow 2\times \large\frac{2\tan \beta}{1+\tan^2\beta}.\frac{1-\tan^2\beta}{1+\tan^2\beta}$
$\Rightarrow 2\times \large\frac{2\times (1/3)}{1+(1/9)}.\frac{1-(1/9)}{1+(1/9)}$
$\Rightarrow 2\times \large\frac{\Large\frac{2}{3}}{\Large\frac{10}{9}}\times \large\frac{\Large\frac{8}{9}}{\Large\frac{10}{9}}$
$\Rightarrow 2\times \large\frac{2}{3}\times \frac{9}{10}\times \frac{8}{9}\times \frac{9}{10}$
$\Rightarrow \large\frac{24}{25}$
Hence $\cos 2\alpha=\sin 4\beta$
Hence (B) is the correct answer.
answered Apr 21, 2014 by sreemathi.v
 

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