If tanh x = frac{1}{2},then sinh 2x - text{se | vedclass

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(B) Given,$	anh x = \frac{1}{2}$.We know that $	anh^2 x = 1 - 	ext{sech}^2 x$,so $	ext{sech}^2 x = 1 - (\frac{1}{2})^2 = 1 - \frac{1}{4} = \frac{3

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$\frac{11}{15}$

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(B) Given,$	anh x = \frac{1}{2}$.We know that $	anh^2 x = 1 - 	ext{sech}^2 x$,so $	ext{sech}^2 x = 1 - (\frac{1}{2})^2 = 1 - \frac{1}{4} = \frac{3}{4}$.Thus,$	ext{sech } x = \frac{\sqrt{3}}{2}$ and $\cosh x = \frac{2}{\sqrt{3}}$.Since $	anh x = \frac{\sinh x}{\cosh x}$,we have $\sinh x = 	anh x \cdot \cosh x = \frac{1}{2} \cdot \frac{2}{\sqrt{3}} = \frac{1}{\sqrt{3}}$.Now,$\sinh 2x = 2 \sinh x \cosh x = 2 \cdot \frac{1}{\sqrt{3}} \cdot \frac{2}{\sqrt{3}} = \frac{4}{3}$.And $	ext{sech } 2x = \frac{1}{\cosh 2x} = \frac{1}{\cosh^2 x + \sinh^2 x} = \frac{1}{(\frac{2}{\sqrt{3}})^2 + (\frac{1}{\sqrt{3}})^2} = \frac{1}{\frac{4}{3} + \frac{1}{3}} = \frac{1}{\frac{5}{3}} = \frac{3}{5}$.Therefore,$\sinh 2x - 	ext{sech } 2x = \frac{4}{3} - \frac{3}{5} = \frac{20 - 9}{15} = \frac{11}{15}$.

If $\tanh x = \frac{1}{2}$,then $\sinh 2x - \text{sech } 2x = $

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