If alpha ,,beta ,,gamma and delta &nbsp | vedclass

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Question

We have.

$	an \left(	heta+\frac{\pi}{4}ight)=3 	an 3 	heta$

$\Rightarrow \frac{1+	an 	heta}{1-	an 	heta}=3 \cdot \frac{3 	an 	heta-\

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Vedclass · Answer

We have.

$	an \left(	heta+\frac{\pi}{4}ight)=3 	an 3 	heta$

$\Rightarrow \frac{1+	an 	heta}{1-	an 	heta}=3 \cdot \frac{3 	an 	heta-	an ^{3} 	heta}{1-3 	an ^{2} 	heta}$

$\Rightarrow \frac{1+t}{1-t}=3\left(\frac{3 t-t^{3}}{1-3 t^{2}}ight) \quad[	ext { putting } t=	an 	heta]$

$\Rightarrow 3 t^{4}-6 t^{2}+8 t-1=0$

$\mathrm{So}, \mathrm{t}_{1}+\mathrm{t}_{2}+\mathrm{t}_{3}+\mathrm{t}_{4}=\frac{0}{3}=0$

$	herefore $ $	an \alpha+	an \beta+	an \gamma+	an \delta=0$

Hence, $( 4)$ is the correct answer.

If $\alpha ,\,\beta ,\,\gamma $ and $\delta $ are the solutions of the equation $\tan \left( {\theta + \frac{\pi }{4}} \right) = 3\,\tan \,3\theta $ , no two of which have equal tangents, then the value of $tan\, \alpha + tan\, \beta + tan\, \gamma + tan\, \delta $ is

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