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Home  >>  CBSE XI  >>  Math  >>  Trigonometric Functions
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$\tan A=\large\frac{1}{2}$$,\tan B=\large\frac{1}{3}$,then $\tan (2A+B)$ is equal to

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

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1 Answer

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Toolbox:
  • $\tan (A+B)=\large\frac{\tan A+\tan B}{1-\tan A\tan B}$
  • $\tan 2A=\large\frac{2\tan A}{1-\tan^2A}$
$\tan (2A+B)=\large\frac{\tan 2A+\tan B}{1-\tan 2A.\tan B}$
$\Rightarrow \large\frac{\Large\frac{2\tan A}{1-\tan^2A}+\frac{1}{3}}{1-\big(\Large\frac{2\tan A}{1-\tan^2A}\big).\frac{1}{3}}$
$\Rightarrow \large\frac{\Large\frac{2\times (1/2)}{1-(1/4)}+\frac{1}{3}}{1-\big(\Large\frac{2\times (1/2)}{1-(1/4)}\times \frac{1}{3}}$
$\Rightarrow \large\frac{\Large\frac{4}{3}+\frac{1}{3}}{1-\Large\frac{4}{9}}$
$\Rightarrow \large\frac{\Large\frac{5}{3}}{\Large\frac{5}{4}}$
$\Rightarrow \large\frac{5}{3}\times \frac{9}{5}$
$\Rightarrow 3$
Hence (C) is the correct answer.
answered Apr 21, 2014 by sreemathi.v
 

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