If theta = frac{pi}{9},then 1 + 27 tan^2 thet | vedclass

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(B) We know that for $	heta = \frac{\pi}{9}$,$3	heta = \frac{\pi}{3}$.$	an(3	heta) = 	an(\frac{\pi}{3}) = \sqrt{3}$.Using the formula $	an(3	he

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$4$

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(B) We know that for $	heta = \frac{\pi}{9}$,$3	heta = \frac{\pi}{3}$.$	an(3	heta) = 	an(\frac{\pi}{3}) = \sqrt{3}$.Using the formula $	an(3	heta) = \frac{3	an	heta - 	an^3	heta}{1 - 3	an^2	heta}$,we have:$\frac{3	an	heta - 	an^3	heta}{1 - 3	an^2	heta} = \sqrt{3}$.Squaring both sides:$(3	an	heta - 	an^3	heta)^2 = 3(1 - 3	an^2	heta)^2$.$9	an^2	heta + 	an^6	heta - 6	an^4	heta = 3(1 + 9	an^4	heta - 6	an^2	heta)$.$9	an^2	heta + 	an^6	heta - 6	an^4	heta = 3 + 27	an^4	heta - 18	an^2	heta$.Rearranging the terms:$	an^6	heta - 33	an^4	heta + 27	an^2	heta + 1 = 4$.

If $\theta = \frac{\pi}{9}$,then $1 + 27 \tan^2 \theta - 33 \tan^4 \theta + \tan^6 \theta = $

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