If operatorname{Sinh}^{-1} x = log 3 and oper | vedclass

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(A) Given $\operatorname{Sinh}^{-1} x = \log 3$. Since $\operatorname{Sinh}^{-1} x = \ln(x + \sqrt{x^2 + 1})$,we have $x + \sqrt{x^2 + 1} = 3$. Let $x

Vedclass · Accepted Answer

$\log \sqrt{\frac{5}{3}}$

Vedclass · Answer

(A) Given $\operatorname{Sinh}^{-1} x = \log 3$. Since $\operatorname{Sinh}^{-1} x = \ln(x + \sqrt{x^2 + 1})$,we have $x + \sqrt{x^2 + 1} = 3$. Let $x + \sqrt{x^2 + 1} = e^{\log 3} = 3$. Then $\sqrt{x^2 + 1} = 3 - x$. Squaring both sides: $x^2 + 1 = 9 - 6x + x^2$,which gives $6x = 8$,so $x = \frac{4}{3}$. Given $\operatorname{Cosh}^{-1} y = \log \frac{3}{2}$. Since $\operatorname{Cosh}^{-1} y = \ln(y + \sqrt{y^2 - 1})$,we have $y + \sqrt{y^2 - 1} = \frac{3}{2}$. Then $\sqrt{y^2 - 1} = \frac{3}{2} - y$. Squaring both sides: $y^2 - 1 = \frac{9}{4} - 3y + y^2$,which gives $3y = \frac{9}{4} + 1 = \frac{13}{4}$,so $y = \frac{13}{12}$. Now,$x - y = \frac{4}{3} - \frac{13}{12} = \frac{16 - 13}{12} = \frac{3}{12} = \frac{1}{4}$. We need to find $\operatorname{Tanh}^{-1}(\frac{1}{4})$. Using the formula $\operatorname{Tanh}^{-1} z = \frac{1}{2} \log \left( \frac{1+z}{1-z} \right)$,we get: $\operatorname{Tanh}^{-1}(\frac{1}{4}) = \frac{1}{2} \log \left( \frac{1 + 1/4}{1 - 1/4} \right) = \frac{1}{2} \log \left( \frac{5/4}{3/4} \right) = \frac{1}{2} \log \left( \frac{5}{3} \right) = \log \sqrt{\frac{5}{3}}$.

If $\operatorname{Sinh}^{-1} x = \log 3$ and $\operatorname{Cosh}^{-1} y = \log \frac{3}{2}$,then $\operatorname{Tanh}^{-1}(x-y) = $

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