frac{sinh(x+y) + sinh(x-y)}{cosh(x+y) - cosh( | vedclass

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Question

(B) We use the sum and difference formulas for hyperbolic functions:$\sinh(x+y) = \sinh x \cosh y + \cosh x \sinh y$$\sinh(x-y) = \sinh x \cosh y - \c

Vedclass · Accepted Answer

$\coth y$

Vedclass · Answer

(B) We use the sum and difference formulas for hyperbolic functions:$\sinh(x+y) = \sinh x \cosh y + \cosh x \sinh y$$\sinh(x-y) = \sinh x \cosh y - \cosh x \sinh y$$\cosh(x+y) = \cosh x \cosh y + \sinh x \sinh y$$\cosh(x-y) = \cosh x \cosh y - \sinh x \sinh y$Substituting these into the numerator:$\sinh(x+y) + \sinh(x-y) = (\sinh x \cosh y + \cosh x \sinh y) + (\sinh x \cosh y - \cosh x \sinh y) = 2 \sinh x \cosh y$Substituting these into the denominator:$\cosh(x+y) - \cosh(x-y) = (\cosh x \cosh y + \sinh x \sinh y) - (\cosh x \cosh y - \sinh x \sinh y) = 2 \sinh x \sinh y$Now,divide the numerator by the denominator:$\frac{2 \sinh x \cosh y}{2 \sinh x \sinh y} = \frac{\cosh y}{\sinh y} = \coth y$Therefore,the correct option is $B$.

$\frac{\sinh(x+y) + \sinh(x-y)}{\cosh(x+y) - \cosh(x-y)} = $

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