For theta in left(0, frac{pi}{2}right),if tan | vedclass

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(A) We know that $	an 3	heta = 	an(2	heta + 	heta)$.Using the formula $	an(A + B) = \frac{	an A + 	an B}{1 - 	an A 	an B}$,we get:$	an 3	h

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$\frac{\pi}{12}$

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(A) We know that $	an 3	heta = 	an(2	heta + 	heta)$.Using the formula $	an(A + B) = \frac{	an A + 	an B}{1 - 	an A 	an B}$,we get:$	an 3	heta = \frac{	an 2	heta + 	an 	heta}{1 - 	an 2	heta 	an 	heta}$.Rearranging this,we have:$	an 3	heta(1 - 	an 2	heta 	an 	heta) = 	an 2	heta + 	an 	heta$.$	an 3	heta - 	an 3	heta 	an 2	heta 	an 	heta = 	an 2	heta + 	an 	heta$.$	an 3	heta 	an 2	heta 	an 	heta = 	an 3	heta - 	an 2	heta - 	an 	heta$.Given the equation $	an 3	heta 	an 2	heta 	an 	heta + 	an 2	heta + 	an 	heta = 1$,substitute the expression derived above:$(	an 3	heta - 	an 2	heta - 	an 	heta) + 	an 2	heta + 	an 	heta = 1$.$	an 3	heta = 1$.Since $	an 3	heta = 	an \frac{\pi}{4}$,we have $3	heta = \frac{\pi}{4} + n\pi$.For $	heta \in \left(0, \frac{\pi}{2}ight)$,we take $n = 0$,so $3	heta = \frac{\pi}{4}$,which gives $	heta = \frac{\pi}{12}$.

For $\theta \in \left(0, \frac{\pi}{2}\right)$,if $\tan 3\theta \cdot \tan 2\theta \cdot \tan \theta + \tan 2\theta + \tan \theta = 1$,then $\theta =$

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