The general solution of tan theta + tan 2thet | vedclass

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(B) Given equation: $	an 	heta + 	an 2	heta = 	an 3	heta$. Using the identity $	an 3	heta = 	an(2	heta + 	heta) = \frac{	an 2	heta + 	an

Vedclass · Accepted Answer

$	heta = n\pi, n \in \mathbb{Z}$ or $	heta = \frac{p\pi}{3}, p \in \mathbb{Z}$

Vedclass · Answer

(B) Given equation: $	an 	heta + 	an 2	heta = 	an 3	heta$. Using the identity $	an 3	heta = 	an(2	heta + 	heta) = \frac{	an 2	heta + 	an 	heta}{1 - 	an 2	heta 	an 	heta}$. Substituting this into the given equation: $	an 	heta + 	an 2	heta = \frac{	an 2	heta + 	an 	heta}{1 - 	an 2	heta 	an 	heta}$. This implies $(	an 	heta + 	an 2	heta) \left(1 - \frac{1}{1 - 	an 2	heta 	an 	heta}ight) = 0$. Case $1$: $	an 	heta + 	an 2	heta = 0$ $\Rightarrow \frac{\sin 3	heta}{\cos 	heta \cos 2	heta} = 0$ $\Rightarrow 3	heta = n\pi$ $\Rightarrow 	heta = \frac{n\pi}{3}$. Case $2$: $1 - 	an 2	heta 	an 	heta = 1 \Rightarrow 	an 2	heta 	an 	heta = 0$. This implies $	an 	heta = 0$ or $	an 2	heta = 0$. If $	an 	heta = 0$,then $	heta = n\pi$. If $	an 2	heta = 0$,then $2	heta = k\pi \Rightarrow 	heta = \frac{k\pi}{2}$. Combining these,the general solution is $	heta = \frac{n\pi}{3}$ or $	heta = n\pi$. However,checking the domain,$	an 	heta, 	an 2	heta, 	an 3	heta$ must be defined. Thus,$	heta 
eq \frac{(2n+1)\pi}{2}, \frac{(2n+1)\pi}{4}, \frac{(2n+1)\pi}{6}$. The valid solutions are $	heta = n\pi$ or $	heta = \frac{p\pi}{3}$.

The general solution of $\tan \theta + \tan 2\theta = \tan 3\theta$ is

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