If operatorname{Sinh}^{-1}(2)+operatorname{Si | vedclass

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(D) Given that $\operatorname{Sinh}^{-1}(2)+\operatorname{Sinh}^{-1}(3)=\alpha$. Taking $\sinh$ on both sides,we have $\sinh(\operatorname{Sinh}^{-1}(

Vedclass · Accepted Answer

$2 \sqrt{10}+3 \sqrt{5}$

Vedclass · Answer

(D) Given that $\operatorname{Sinh}^{-1}(2)+\operatorname{Sinh}^{-1}(3)=\alpha$. Taking $\sinh$ on both sides,we have $\sinh(\operatorname{Sinh}^{-1}(2)+\operatorname{Sinh}^{-1}(3))=\sinh \alpha$. Let $x=\operatorname{Sinh}^{-1}(2)$ and $y=\operatorname{Sinh}^{-1}(3)$,so $\sinh x=2$ and $\sinh y=3$. We know that $\cosh^2 x - \sinh^2 x = 1$,so $\cosh x = \sqrt{1+\sinh^2 x} = \sqrt{1+2^2} = \sqrt{5}$. Similarly,$\cosh y = \sqrt{1+\sinh^2 y} = \sqrt{1+3^2} = \sqrt{10}$. Using the identity $\sinh(x+y) = \sinh x \cosh y + \cosh x \sinh y$,we get: $\sinh \alpha = \sinh x \cosh y + \cosh x \sinh y = (2)(\sqrt{10}) + (\sqrt{5})(3) = 2 \sqrt{10} + 3 \sqrt{5}$.

If $\operatorname{Sinh}^{-1}(2)+\operatorname{Sinh}^{-1}(3)=\alpha$,then $\sinh \alpha=$

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