If tanh(x) = frac{1}{3},then tanh(3x) is | vedclass

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(B) We are given $	anh(x) = \frac{1}{3}$.Using the formula $	anh(3x) = \frac{3	anh(x) + 	anh^3(x)}{1 + 3	anh^2(x)}$:Substitute $	anh(x) = \frac{

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$\frac{7}{9}$

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(B) We are given $	anh(x) = \frac{1}{3}$.Using the formula $	anh(3x) = \frac{3	anh(x) + 	anh^3(x)}{1 + 3	anh^2(x)}$:Substitute $	anh(x) = \frac{1}{3}$ into the formula:$	anh(3x) = \frac{3(\frac{1}{3}) + (\frac{1}{3})^3}{1 + 3(\frac{1}{3})^2}$$	anh(3x) = \frac{1 + \frac{1}{27}}{1 + 3(\frac{1}{9})}$$	anh(3x) = \frac{\frac{28}{27}}{1 + \frac{1}{3}}$$	anh(3x) = \frac{\frac{28}{27}}{\frac{4}{3}}$$	anh(3x) = \frac{28}{27} 	imes \frac{3}{4} = \frac{7}{9}$

If $\tanh(x) = \frac{1}{3}$,then $\tanh(3x)$ is

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