The solution of tan 2theta tan theta = 1 is | vedclass

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(B) Given $	an 2	heta 	an 	heta = 1$.Using the identity $	an 2	heta = \frac{2 	an 	heta}{1 - 	an^2 	heta}$,we have:$\frac{2 	an 	heta}{1 -

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$(6n \pm 1)\frac{\pi}{6}$

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(B) Given $	an 2	heta 	an 	heta = 1$.Using the identity $	an 2	heta = \frac{2 	an 	heta}{1 - 	an^2 	heta}$,we have:$\frac{2 	an 	heta}{1 - 	an^2 	heta} \cdot 	an 	heta = 1$$2 	an^2 	heta = 1 - 	an^2 	heta$$3 	an^2 	heta = 1$$	an^2 	heta = \frac{1}{3} \Rightarrow 	an 	heta = \pm \frac{1}{\sqrt{3}}$This implies $	heta = n\pi \pm \frac{\pi}{6} = \frac{6n\pi \pm \pi}{6} = (6n \pm 1)\frac{\pi}{6}$ for $n \in \mathbb{Z}$.

The solution of $\tan 2\theta \tan \theta = 1$ is

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