sinh[log(2+sqrt{5})] + cosh[log(2+sqrt{3})] = | vedclass

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(A) We know that $\sinh(x) = \frac{e^x - e^{-x}}{2}$ and $\cosh(x) = \frac{e^x + e^{-x}}{2}$.For the first term: $\sinh[\log(2+\sqrt{5})] = \frac{e^{\

Vedclass · Accepted Answer

$4$

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(A) We know that $\sinh(x) = \frac{e^x - e^{-x}}{2}$ and $\cosh(x) = \frac{e^x + e^{-x}}{2}$.For the first term: $\sinh[\log(2+\sqrt{5})] = \frac{e^{\log(2+\sqrt{5})} - e^{-\log(2+\sqrt{5})}}{2} = \frac{(2+\sqrt{5}) - \frac{1}{2+\sqrt{5}}}{2}$.Since $\frac{1}{2+\sqrt{5}} = \sqrt{5}-2$,we have $\frac{(2+\sqrt{5}) - (\sqrt{5}-2)}{2} = \frac{4}{2} = 2$.For the second term: $\cosh[\log(2+\sqrt{3})] = \frac{e^{\log(2+\sqrt{3})} + e^{-\log(2+\sqrt{3})}}{2} = \frac{(2+\sqrt{3}) + \frac{1}{2+\sqrt{3}}}{2}$.Since $\frac{1}{2+\sqrt{3}} = 2-\sqrt{3}$,we have $\frac{(2+\sqrt{3}) + (2-\sqrt{3})}{2} = \frac{4}{2} = 2$.Adding these results: $2 + 2 = 4$.

$\sinh[\log(2+\sqrt{5})] + \cosh[\log(2+\sqrt{3})] = ?$

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