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If $\tan\alpha=\large\frac{m}{m+1}$$,\tan \beta=\large\frac{1}{2m+1}$,then $\alpha+\beta$ is equal to

$\begin{array}{1 1}(A)\;\large\frac{\pi}{2}&(B)\;\large\frac{\pi}{3}\\(C)\;\large\frac{\pi}{6}&(D)\;\large\frac{\pi}{4}\end{array} $

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1 Answer

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  • $\tan (\alpha+\beta)=\large\frac{\tan \alpha+\tan \beta}{1-\tan \alpha\tan\beta}$
Given :
$\tan\alpha=\large\frac{m}{m+1},$$\tan \beta=\large\frac{1}{2m+1}$
$\tan(\alpha+\beta)=\large\frac{\tan\alpha+\tan \beta}{1-\tan \alpha.\tan \beta}$
$\Rightarrow \large\frac{\Large\frac{m}{m+1}+\frac{1}{2m+1}}{1-\Large\frac{m}{(m+1)(2m+1)}}$
$\Rightarrow \large\frac{\Large\frac{m(2m+1)+(m+1)}{(m+1)(2m+1)}}{\Large\frac{(m+1)(2m+1)-m}{(m+1)(2m+1)}}$
$\Rightarrow \large\frac{m(2m+1)+m+1}{(m+1)(2m+1)-m}$
$\Rightarrow \large\frac{2m^2+m+m+1}{2m^2+m+2m+1-m}$
$\Rightarrow \tan(\alpha+\beta)=1=\tan\large\frac{\pi}{4}$
$\therefore \alpha+\beta=\large\frac{\pi}{4}$
Hence (D) is the correct answer.
answered Apr 21, 2014 by sreemathi.v
 

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