If A, B, C are angles of a triangle ABC,then | vedclass

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(B) In $	riangle ABC$,$A + B + C = \pi$,so $2A + 2B + 2C = 2\pi$. Therefore,$2A + 2B = 2\pi - 2C$. Taking tangent on both sides: $	an(2A + 2B) = 	a

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$	an 2A 	an 2B 	an 2C$

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(B) In $	riangle ABC$,$A + B + C = \pi$,so $2A + 2B + 2C = 2\pi$. Therefore,$2A + 2B = 2\pi - 2C$. Taking tangent on both sides: $	an(2A + 2B) = 	an(2\pi - 2C) = -	an 2C$. Using the formula $	an(x + y) = \frac{	an x + 	an y}{1 - 	an x 	an y}$,we get: $\frac{	an 2A + 	an 2B}{1 - 	an 2A 	an 2B} = -	an 2C$. Multiplying both sides by $(1 - 	an 2A 	an 2B)$: $	an 2A + 	an 2B = -	an 2C(1 - 	an 2A 	an 2B)$. $	an 2A + 	an 2B = -	an 2C + 	an 2A 	an 2B 	an 2C$. Rearranging the terms: $	an 2A + 	an 2B + 	an 2C = 	an 2A 	an 2B 	an 2C$.

List-$I$	List-$II$
$(I) \ 3 \sin^2 x + 4 \cos^2 x - 2$	$(a) \ [\frac{1}{4}, 1]$
$(II) \ \cos^2 x + \sin^4 x$	$(b) \ [-\frac{1}{4}, \frac{1}{4}]$
$(III) \ \sin^6 x + \cos^6 x$	$(c) \ [1, 2]$
$(IV) \ \cos x \cos(\frac{2 \pi}{3} + x) \cos(\frac{2 \pi}{3} - x)$	$(d) \ [\frac{3}{4}, 1]$
	$(e) \ [0, 1]$

If $A, B, C$ are angles of a $\triangle ABC$,then $\tan 2A + \tan 2B + \tan 2C =$

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