If sinh x = frac{3}{4} and cosh y = frac{5}{3 | vedclass

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(B) Given $\sinh x = \frac{3}{4}$.Using the formula $\sinh^{-1} z = \log(z + \sqrt{z^2 + 1})$:$x = \log(\frac{3}{4} + \sqrt{(\frac{3}{4})^2 + 1}) = \l

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$\log 6$

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(B) Given $\sinh x = \frac{3}{4}$.Using the formula $\sinh^{-1} z = \log(z + \sqrt{z^2 + 1})$:$x = \log(\frac{3}{4} + \sqrt{(\frac{3}{4})^2 + 1}) = \log(\frac{3}{4} + \sqrt{\frac{9}{16} + 1}) = \log(\frac{3}{4} + \frac{5}{4}) = \log(2)$.Given $\cosh y = \frac{5}{3}$.Using the formula $\cosh^{-1} z = \log(z + \sqrt{z^2 - 1})$:$y = \log(\frac{5}{3} + \sqrt{(\frac{5}{3})^2 - 1}) = \log(\frac{5}{3} + \sqrt{\frac{25}{9} - 1}) = \log(\frac{5}{3} + \frac{4}{3}) = \log(3)$.Therefore,$x + y = \log 2 + \log 3 = \log(2 	imes 3) = \log 6$.

If $\sinh x = \frac{3}{4}$ and $\cosh y = \frac{5}{3}$,then $x + y =$

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