If sinh ^{-1}(sqrt{8})+sinh ^{-1}(sqrt{24})=a | vedclass

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(B) We have, $\sinh ^{-1}(\sqrt{8})+\sinh ^{-1}(\sqrt{24})=\alpha$. Let $\sinh ^{-1}(\sqrt{8})=x$, then $\sinh x = \sqrt{8}$. Since $\cosh^2 x - \sinh

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$6 \sqrt{6}+10 \sqrt{2}$

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(B) We have, $\sinh ^{-1}(\sqrt{8})+\sinh ^{-1}(\sqrt{24})=\alpha$. Let $\sinh ^{-1}(\sqrt{8})=x$, then $\sinh x = \sqrt{8}$. Since $\cosh^2 x - \sinh^2 x = 1$, we have $\cosh x = \sqrt{1 + \sinh^2 x} = \sqrt{1 + 8} = \sqrt{9} = 3$. Let $\sinh ^{-1}(\sqrt{24})=y$, then $\sinh y = \sqrt{24}$. Similarly, $\cosh y = \sqrt{1 + \sinh^2 y} = \sqrt{1 + 24} = \sqrt{25} = 5$. Using the identity $\sinh(x+y) = \sinh x \cosh y + \cosh x \sinh y$, we get: $\sinh(x+y) = (\sqrt{8})(5) + (3)(\sqrt{24}) = 5(2\sqrt{2}) + 3(2\sqrt{6}) = 10\sqrt{2} + 6\sqrt{6}$. Since $\alpha = x+y$, we have $\sinh \alpha = 6\sqrt{6} + 10\sqrt{2}$.

If $\sinh ^{-1}(\sqrt{8})+\sinh ^{-1}(\sqrt{24})=\alpha$, then $\sinh \alpha=$

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