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

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Question

(C) Given $\sinh x = -\frac{1}{2}$.We know the identity $\cosh^2 x - \sinh^2 x = 1$.Substituting the value: $\cosh^2 x = 1 + (-\frac{1}{2})^2 = 1 + \f

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$-\frac{\sqrt{5}}{3}$

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(C) Given $\sinh x = -\frac{1}{2}$.We know the identity $\cosh^2 x - \sinh^2 x = 1$.Substituting the value: $\cosh^2 x = 1 + (-\frac{1}{2})^2 = 1 + \frac{1}{4} = \frac{5}{4}$.Since $\cosh x > 0$,we have $\cosh x = \frac{\sqrt{5}}{2}$.Now,$	anh x = \frac{\sinh x}{\cosh x} = \frac{-1/2}{\sqrt{5}/2} = -\frac{1}{\sqrt{5}}$.Using the double angle formula $	anh 2x = \frac{2 	anh x}{1 + 	anh^2 x}$:$	anh 2x = \frac{2(-1/\sqrt{5})}{1 + (-1/\sqrt{5})^2} = \frac{-2/\sqrt{5}}{1 + 1/5} = \frac{-2/\sqrt{5}}{6/5} = -\frac{2}{\sqrt{5}} 	imes \frac{5}{6} = -\frac{\sqrt{5}}{3}$.

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

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