If operatorname{Tanh}^{-1} x = operatorname{C | vedclass

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Question

(A) Given $\operatorname{Tanh}^{-1} x = \log \sqrt{5}$.Using the definition $\operatorname{Tanh}^{-1} x = \frac{1}{2} \log \left( \frac{1+x}{1-x} \rig

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$\frac{\pi}{4}$

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(A) Given $\operatorname{Tanh}^{-1} x = \log \sqrt{5}$.Using the definition $\operatorname{Tanh}^{-1} x = \frac{1}{2} \log \left( \frac{1+x}{1-x} \right)$,we have $\frac{1}{2} \log \left( \frac{1+x}{1-x} \right) = \log \sqrt{5} = \frac{1}{2} \log 5$.Thus,$\frac{1+x}{1-x} = 5 \implies 1+x = 5-5x \implies 6x = 4 \implies x = \frac{2}{3}$.Given $\operatorname{Coth}^{-1} y = \log \sqrt{5}$.Using the definition $\operatorname{Coth}^{-1} y = \frac{1}{2} \log \left( \frac{y+1}{y-1} \right)$,we have $\frac{1}{2} \log \left( \frac{y+1}{y-1} \right) = \frac{1}{2} \log 5$.Thus,$\frac{y+1}{y-1} = 5 \implies y+1 = 5y-5 \implies 4y = 6 \implies y = \frac{3}{2}$.Now,$xy = \left( \frac{2}{3} \right) \left( \frac{3}{2} \right) = 1$.Therefore,$\operatorname{Tan}^{-1}(xy) = \operatorname{Tan}^{-1}(1) = \frac{\pi}{4}$.

If $\operatorname{Tanh}^{-1} x = \operatorname{Coth}^{-1} y = \log \sqrt{5}$,then $\operatorname{Tan}^{-1}(xy) = $

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