In triangle ABC,frac{tan A}{2} = frac{tan B}{ | vedclass

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(B) In a $	riangle ABC$,we are given that $\frac{	an A}{2} = \frac{	an B}{3} = \frac{	an C}{4} = k$. Thus,$	an A = 2k$,$	an B = 3k$,and $	an C

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$\frac{111}{8}$

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(B) In a $	riangle ABC$,we are given that $\frac{	an A}{2} = \frac{	an B}{3} = \frac{	an C}{4} = k$. Thus,$	an A = 2k$,$	an B = 3k$,and $	an C = 4k$. In any $	riangle ABC$,the sum of angles is $180^\circ$,which implies $	an A + 	an B + 	an C = 	an A 	an B 	an C$. Substituting the values,we get $2k + 3k + 4k = (2k)(3k)(4k)$,which simplifies to $9k = 24k^3$. Since $k 
eq 0$,we have $k^2 = \frac{9}{24} = \frac{3}{8}$. We need to find $\sec^2 A + \sec^2 B + \sec^2 C = (1 + 	an^2 A) + (1 + 	an^2 B) + (1 + 	an^2 C) = 3 + 	an^2 A + 	an^2 B + 	an^2 C$. Substituting the values,we get $3 + (2k)^2 + (3k)^2 + (4k)^2 = 3 + k^2(4 + 9 + 16) = 3 + 29k^2$. Substituting $k^2 = \frac{3}{8}$,we get $3 + 29 	imes \frac{3}{8} = 3 + \frac{87}{8} = \frac{24 + 87}{8} = \frac{111}{8}$.

In triangle $ABC$,$\frac{\tan A}{2} = \frac{\tan B}{3} = \frac{\tan C}{4}$,then the value of $\sec^2 A + \sec^2 B + \sec^2 C$ is:

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