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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Trignometry
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If $\alpha,\beta,\gamma,\delta$ are the solutions of the equation $\tan\big(\theta+\large\frac{\pi}{4}\big)=$$3\tan 3\theta$,no two of which have equal tangents,then the value of $\tan\alpha+\tan\beta+\tan\gamma+\tan\delta$ is equal to

$(a)\;1\qquad(b)\;-1\qquad(c)\;2\qquad(d)\;0$

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1 Answer

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We have $\tan\big(\theta+\large\frac{\pi}{4}\big)$$=3\tan\theta$
$\large\frac{1+\tan\theta}{1-\tan\theta}$$=3$
$\Rightarrow \large\frac{3\tan\theta-\tan^3\theta}{1-3\tan^2\theta}$
$\large\frac{1+t}{1-t}$$=3\big[\large\frac{3t-t^3}{1-3t^2}\big]$
Put $t=\tan\theta$
$3t^4-6t^2+8t-1=0$
Hence $t_1+t_2+t_3+t_4=\large\frac{0}{3}$$=0$
(or) $\tan\alpha+\tan\beta+\tan\gamma+\tan\delta=0$
Hence (d) is the correct option.
answered Oct 15, 2013 by sreemathi.v
 

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