If sinh x = tan A,then |tanh x| = | vedclass

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(A) Given $\sinh x = 	an A$. We know that $\sinh x = \frac{e^x - e^{-x}}{2} = 	an A$,so $e^x - e^{-x} = 2 	an A$. Let $e^x = t$. Then $t - \frac{1}

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$|\sin A|$

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(A) Given $\sinh x = 	an A$. We know that $\sinh x = \frac{e^x - e^{-x}}{2} = 	an A$,so $e^x - e^{-x} = 2 	an A$. Let $e^x = t$. Then $t - \frac{1}{t} = 2 	an A$,which implies $t^2 - 2 	an A \cdot t - 1 = 0$. Solving for $t$ using the quadratic formula: $t = \frac{2 	an A \pm \sqrt{4 	an^2 A + 4}}{2} = 	an A \pm \sec A$. Since $e^x > 0$,we take $e^x = 	an A + \sec A$. Then $e^{-x} = \frac{1}{\sec A + 	an A} = \sec A - 	an A$. Now,$	anh x = \frac{e^x - e^{-x}}{e^x + e^{-x}} = \frac{(	an A + \sec A) - (\sec A - 	an A)}{(	an A + \sec A) + (\sec A - 	an A)} = \frac{2 	an A}{2 \sec A} = \frac{	an A}{\sec A} = \sin A$. Thus,$|	anh x| = |\sin A|$.

If $\sinh x = \tan A$,then $|\tanh x| =$

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