If cosh ^{-1}left(frac{5}{3}right)+sinh ^{-1} | vedclass

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(C) We use the logarithmic forms of inverse hyperbolic functions: $\cosh ^{-1}(x) = \ln(x + \sqrt{x^2 - 1})$ and $\sinh ^{-1}(x) = \ln(x + \sqrt{x^2 +

Vedclass · Accepted Answer

$6$

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(C) We use the logarithmic forms of inverse hyperbolic functions: $\cosh ^{-1}(x) = \ln(x + \sqrt{x^2 - 1})$ and $\sinh ^{-1}(x) = \ln(x + \sqrt{x^2 + 1})$.Given $\cosh ^{-1}\left(\frac{5}{3}ight) + \sinh ^{-1}\left(\frac{3}{4}ight) = k$.Substituting the logarithmic forms:$\ln\left(\frac{5}{3} + \sqrt{\left(\frac{5}{3}ight)^2 - 1}ight) + \ln\left(\frac{3}{4} + \sqrt{\left(\frac{3}{4}ight)^2 + 1}ight) = k$$\ln\left(\frac{5}{3} + \sqrt{\frac{25}{9} - 1}ight) + \ln\left(\frac{3}{4} + \sqrt{\frac{9}{16} + 1}ight) = k$$\ln\left(\frac{5}{3} + \sqrt{\frac{16}{9}}ight) + \ln\left(\frac{3}{4} + \sqrt{\frac{25}{16}}ight) = k$$\ln\left(\frac{5}{3} + \frac{4}{3}ight) + \ln\left(\frac{3}{4} + \frac{5}{4}ight) = k$$\ln\left(\frac{9}{3}ight) + \ln\left(\frac{8}{4}ight) = k$$\ln(3) + \ln(2) = k$$\ln(3 	imes 2) = k$$\ln(6) = k$Therefore, $e^k = 6$.

If $\cosh ^{-1}\left(\frac{5}{3}\right)+\sinh ^{-1}\left(\frac{3}{4}\right)=k$, then $e^k=$

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