If cosh x = frac{5}{4},then tanh 3x = | vedclass

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(A) Given,$\cosh x = \frac{5}{4}$.We know that $\cosh x = \frac{e^x + e^{-x}}{2}$,so $e^x + e^{-x} = \frac{5}{2}$.Using the identity $\cosh^2 x - \sin

Vedclass · Accepted Answer

$\frac{63}{65}$

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(A) Given,$\cosh x = \frac{5}{4}$.We know that $\cosh x = \frac{e^x + e^{-x}}{2}$,so $e^x + e^{-x} = \frac{5}{2}$.Using the identity $\cosh^2 x - \sinh^2 x = 1$,we get $\sinh^2 x = (\frac{5}{4})^2 - 1 = \frac{25}{16} - 1 = \frac{9}{16}$.Thus,$\sinh x = \frac{3}{4}$ (assuming $x > 0$),which means $\frac{e^x - e^{-x}}{2} = \frac{3}{4}$,so $e^x - e^{-x} = \frac{3}{2}$.We use the formula $	anh 3x = \frac{3 	anh x + 	anh^3 x}{1 + 3 	anh^2 x}$.First,find $	anh x = \frac{\sinh x}{\cosh x} = \frac{3/4}{5/4} = \frac{3}{5}$.Now,$	anh 3x = \frac{3(\frac{3}{5}) + (\frac{3}{5})^3}{1 + 3(\frac{3}{5})^2} = \frac{\frac{9}{5} + \frac{27}{125}}{1 + 3(\frac{9}{25})} = \frac{\frac{225 + 27}{125}}{\frac{25 + 27}{25}} = \frac{252}{125} 	imes \frac{25}{52} = \frac{252}{5 	imes 52} = \frac{252}{260} = \frac{63}{65}$.

If $\cosh x = \frac{5}{4}$,then $\tanh 3x =$

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