If $\sinh (x + y) \cosh (x - y) = k (\sinh 2 x + \sinh 2 y)$ then $k =$

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$\frac{1}{2}$

$- \frac{1}{2}$

answer is B.

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Detailed Solution

$\sinh (x + y) \cosh (x - y)$
$= [\frac{e^{x - y} + e^{- (x - y)}}{2}] [\frac{e^{x + y} - e^{- (x + y)}}{2}] = \frac{1}{2} [\frac{e^{2 x} - e^{- 2 x}}{2} + \frac{e^{2 y} - e^{- 2 y}}{2}] = \frac{1}{2} [\sinh 2 x + \sinh 2 y]$

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