If alpha, beta, gamma and delta are the solut | vedclass

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(D) Given the equation: $	an \left( 	heta + \frac{\pi}{4} ight) = 3 	an 3	heta$.Using the identity $	an(A+B) = \frac{	an A + 	an B}{1 - 	an

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(D) Given the equation: $	an \left( 	heta + \frac{\pi}{4} ight) = 3 	an 3	heta$.Using the identity $	an(A+B) = \frac{	an A + 	an B}{1 - 	an A 	an B}$ and $	an 3	heta = \frac{3	an 	heta - 	an^3 	heta}{1 - 3	an^2 	heta}$,we get:$\frac{1 + 	an 	heta}{1 - 	an 	heta} = 3 \left( \frac{3	an 	heta - 	an^3 	heta}{1 - 3	an^2 	heta} ight)$.Let $t = 	an 	heta$. Then:$\frac{1 + t}{1 - t} = \frac{9t - 3t^3}{1 - 3t^2}$.Cross-multiplying gives:$(1 + t)(1 - 3t^2) = (1 - t)(9t - 3t^3)$.$1 - 3t^2 + t - 3t^3 = 9t - 3t^3 - 9t^2 + 3t^4$.Rearranging terms to one side:$3t^4 - 6t^2 + 8t - 1 = 0$.This is a quartic equation in $t$ of the form $at^4 + bt^3 + ct^2 + dt + e = 0$,where $a=3, b=0, c=-6, d=8, e=-1$.The sum of the roots $t_1 + t_2 + t_3 + t_4 = -\frac{b}{a} = -\frac{0}{3} = 0$.Since $\alpha, \beta, \gamma, \delta$ are the solutions,$	an \alpha, 	an \beta, 	an \gamma, 	an \delta$ are the roots $t_1, t_2, t_3, t_4$.Therefore,$	an \alpha + 	an \beta + 	an \gamma + 	an \delta = 0$.

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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