The solution of tan,, 2theta,, tantheta = 1 i | vedclass

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Question

$	an 2 	heta 	an 	heta=1 \Rightarrow \frac{2 	an 	heta}{1-	an ^{2} 	heta} \cdot 	an 	heta=1$

$\Rightarrow \quad 2 	an ^{2} 	heta=1-	an

Vedclass · Accepted Answer

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

Vedclass · Answer

$	an 2 	heta 	an 	heta=1 \Rightarrow \frac{2 	an 	heta}{1-	an ^{2} 	heta} \cdot 	an 	heta=1$

$\Rightarrow \quad 2 	an ^{2} 	heta=1-	an ^{2} 	heta \Rightarrow 3 	an ^{2} 	heta=1$

$\Rightarrow \quad 	an 	heta=\pm \frac{1}{\sqrt{3}}=	an \left(\pm \frac{\pi}{6}ight)$

$\Rightarrow \quad 	heta=n \pi \pm \frac{\pi}{6} \quad(m \in Z)$

$=(6 n \pm 1) \frac{\pi}{6}$

$\mathrm{Or}$

$	an 2 	heta=\cot 	heta=	an \left(\frac{\pi}{2}-	hetaight)$

$\Rightarrow \quad 2 	heta=n \pi+\frac{\pi}{2}-	heta$

$\Rightarrow \quad 3 	heta=n \pi+\frac{\pi}{2}$

$\Rightarrow \quad 	heta=\frac{\mathrm{n} \pi}{3}+\frac{\pi}{6}=(2 \mathrm{n}+1) \frac{\pi}{6}$

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

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