2 cosh (x+y) sinh (x-y) + sinh 2y = | vedclass

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(A) We use the identity $2 \cosh A \sinh B = \sinh(A+B) + \sinh(A-B)$.Let $A = x+y$ and $B = x-y$.Then $A+B = (x+y) + (x-y) = 2x$ and $A-B = (x+y) - (

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$\sinh 2x$

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(A) We use the identity $2 \cosh A \sinh B = \sinh(A+B) + \sinh(A-B)$.Let $A = x+y$ and $B = x-y$.Then $A+B = (x+y) + (x-y) = 2x$ and $A-B = (x+y) - (x-y) = 2y$.Substituting these into the expression:$2 \cosh (x+y) \sinh (x-y) = \sinh(2x) + \sinh(2y)$.Now,add $\sinh 2y$ to the expression:$\sinh 2x + \sinh 2y + \sinh 2y = \sinh 2x + 2 \sinh 2y$.Wait,re-evaluating the expression $2 \cosh A \sinh B = \sinh(A+B) - \sinh(A-B)$ is incorrect. The correct identity is $2 \cosh A \sinh B = \sinh(A+B) - \sinh(A-B)$.Let us re-calculate:$2 \cosh (x+y) \sinh (x-y) = \sinh((x+y) + (x-y)) - \sinh((x+y) - (x-y)) = \sinh 2x - \sinh 2y$.Adding $\sinh 2y$ to this result:$(\sinh 2x - \sinh 2y) + \sinh 2y = \sinh 2x$.

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

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