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# If $\alpha,\beta,\gamma,\delta$ are the solutions of the equation $\tan\big(\theta+\large\frac{\pi}{4}\big)=$$3\tan 3\theta,no two of which have equal tangents,then the value of \tan\alpha+\tan\beta+\tan\gamma+\tan\delta is equal to (a)\;1\qquad(b)\;-1\qquad(c)\;2\qquad(d)\;0 Can you answer this question? ## 1 Answer 0 votes We have \tan\big(\theta+\large\frac{\pi}{4}\big)$$=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 $t=\tan\theta$
$3t^4-6t^2+8t-1=0$