sinh ^{-1} 2 + sinh ^{-1} 3 = x Rightarrow co | vedclass

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(C) Given,$\sinh ^{-1} 2 + \sinh ^{-1} 3 = x$. Taking $\cosh$ on both sides,$\cosh x = \cosh(\sinh ^{-1} 2 + \sinh ^{-1} 3)$. Using the identity $\cos

Vedclass · Accepted Answer

$\frac{1}{2}(12 + 2 \sqrt{50})$

Vedclass · Answer

(C) Given,$\sinh ^{-1} 2 + \sinh ^{-1} 3 = x$. Taking $\cosh$ on both sides,$\cosh x = \cosh(\sinh ^{-1} 2 + \sinh ^{-1} 3)$. Using the identity $\cosh(A + B) = \cosh A \cosh B + \sinh A \sinh B$,we get: $\cosh x = \cosh(\sinh ^{-1} 2) \cosh(\sinh ^{-1} 3) + \sinh(\sinh ^{-1} 2) \sinh(\sinh ^{-1} 3)$. Since $\cosh(\sinh ^{-1} y) = \sqrt{1 + y^2}$,we have: $\cosh(\sinh ^{-1} 2) = \sqrt{1 + 2^2} = \sqrt{5}$ and $\cosh(\sinh ^{-1} 3) = \sqrt{1 + 3^2} = \sqrt{10}$. Substituting these values: $\cosh x = (\sqrt{5})(\sqrt{10}) + (2)(3) = \sqrt{50} + 6$. To match the options,we write this as $\frac{2(6 + \sqrt{50})}{2} = \frac{1}{2}(12 + 2 \sqrt{50})$.

$\sinh ^{-1} 2 + \sinh ^{-1} 3 = x \Rightarrow \cosh x$ is equal to

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