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Question Number 51887 by Tawa1 last updated on 31/Dec/18

$$\mathrm{Prove}\mathrm{that};\tanh \left(\log \sqrt{3}\right)=\frac{1}{2}$$

Answered by tanmay.chaudhury50@gmail.com last updated on 01/Jan/19

$$e^{ix}=cosx+isinx$$$$e^{-ix}=cosx-isinx$$$$x=i\theta$$$$e^{-\theta }=cos\left(i\theta \right)+isin\left(i\theta \right)$$$$=cosh\left(\theta \right)-sinh\left(\theta \right)$$$$e^{\theta }=cosh\left(\theta \right)+sinh\left(\theta \right)$$$$tanh\left(\theta \right)=\frac{e^{\theta }-e^{-\theta }}{e^{\theta }+e^{-\theta }}$$$$tanh\left(ln\sqrt{3}\right)=\frac{e^{ln\sqrt{3}}-e^{-ln\sqrt{3}}}{e^{ln\sqrt{3}}+e^{-ln\sqrt{3}}}=\frac{e^{\frac{1}{2}ln3}-e^{-\frac{1}{2}ln3}}{e^{\frac{1}{2}ln3}+e^{-\frac{1}{2}ln3}}.$$$$=\frac{{\left(3\right)}^{\frac{1}{2}}-{\left(3\right)}^{\frac{-1}{2}}}{{\left(3\right)}^{\frac{1}{2}}+{\left(3\right)}^{\frac{-1}{2}}}=\frac{3^{\frac{1}{2}+\frac{1}{2}}-1}{3^{\frac{1}{2}+\frac{1}{2}}+1}.=\frac{3-1}{3+1}=\frac{1}{2}$$$$a^x=e^{xlna}$$

Commented by Tawa1 last updated on 31/Dec/18

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

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