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Relation between circular and hyperbolic function : If tanh$ \; x = \frac{2}{3}$ , find the value of $x$ and then cosh $2x $
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Now, $ tanh \; x = \frac{2}{3} \; \; \therefore \; x \; = \; tanh^{-1}(\frac{2}{3}) \; = \; \frac{1}{2} \; log(\tfrac{1 + \tfrac{2}{3}}{1 - \tfrac{2}{3}}) \; = \; \frac{1}{2} log\frac{3 \; + \; 2}{3 \; - \; 2} \; = \; \frac{1}{2} \; log \; 5 \; = \; log\sqrt{5} \\ \; \\ \; \\ cosh \; 2x = cosh(2 \; log \;\sqrt{5}) \; = \; \frac{ e^{2 \; log \;\sqrt{5}} \; + \; e^{-2 \; log \;\sqrt{5}} }{2} \; = \; \frac{ e^{{ log \;(\sqrt{5})^2}} \; + \; e^{{-2 \; log \;(\sqrt{5})^2}} }{2} \\ \; \\ \; \\ cosh \; 2x = \tfrac{e^{log \;5} \; \; + \; \; e^{- \;log \;5} }{2} \; = \; \tfrac{e^{log \;5} \; \; + \; \; \frac{1}{ e^{log \;5}} }{2} \; = \; \frac{5 \; + \; \frac{1}{5}}{2} \; = \; \frac{13}{5} $

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